homework 3.5 applications of exponentials

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3.2.1: Solving Exponential Equations

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Exponential Equations

When you were first learning equations, you learned the rule that whatever you do to one side of an equation, you must also do to the other side so that the equation stays in balance. The basic techniques of adding, subtracting, multiplying and dividing both sides of an equation worked to solve almost all equations up until now. With logarithms, you have more tools to isolate a variable. Consider the following equation and ask yourself: why is x=3? Logically it makes sense that if the bases match, then the exponents must match as well, but how can it be shown for examples like this one?

1.79898 2x =1.79898 6

Solving Exponential Equations

A common technique for solving equations with unknown variables in exponents is to take the log of the desired base of both sides of the equation. Then, you can use properties of logs to simplify and solve the equation.

Take the following equation. To solve for t, you should first simplify the expression as much as possible and then take the natural log of both sides.

\(\ \begin{aligned} 9,000&=300 \cdot \frac{(1.06)^{t}-1}{0.06} & \\ 30 &=\frac{(1.06)^{t}-1}{0.06} \\ 1.8 &=(1.06)^{t}-1 \\ 2.8 &=1.06^{t} \\ \ln 2.8 &=\ln \left(1.06^{t}\right)=t \cdot \ln (1.06) \\ t &=\frac{\ln 2.8}{\ln 1.06} \approx 17.67 \text { years } \end{aligned}\)

It does not matter what base you use in this situation as long as you use the same base on both sides. Choosing natural log allows you to use a calculator to finish the problem.

Note that this type of equation is common in financial mathematics. The equation above represents the unknown amount of time it will take you to save $9,000 in a savings account if you save $300 at the end of each year in an account that earns 6% annual compound interest.

The other good base to use is base 10. When solving the following equation for x: 16 x =25, you will need to use a calculator to get the final answer and your calculator can handle base 10 as well. First take the log of both sides. Then, use log properties and your calculator to help.

\(\ \begin{aligned} 16^{x} &=25 \\ \log 16^{x} &=\log 25 \\ x \log 16 &=\log 25 \\ x &=\frac{\log 25}{\log 16} \\ x &=1.16 \end{aligned}\)

Earlier, you were asked how to show that if the bases match in an equation, the exponents should match. In the equation, logs can be used to reduce the equation to 2x=6.

Take the log of both sides and use the property of exponentiation of logs to bring the exponent out front.

\(\ \begin{aligned} \log 1.79898^{2 x} &=\log 1.79898^{6} \\ 2 x \cdot \log 1.79898 &=6 \cdot \log 1.79898 \\ 2 x &=6 \\ x &=3 \end{aligned}\)

Solve the following equation for all possible values of x: (x+1) x−4 −1=0

(x+1) x−4 −1=0

(x+1) x−4 =1

Case 1 is that x+1 is positive in which case you can take the log of both sides.

\(\ \begin{array}{c} \log (x+1)^{(x-4)}=\log 1 \\ (x-4) \cdot \log (x+1)=0 \\ x-4=0 \text { or } \log (x+1)=0 \\ x=4 \text { or }(x+1)=10^{0}=1 \\ \quad x=4,0 \end{array}\)

Note that log1=0

Case 2 is that (x+1) is negative 1 and raised to an even power. This happens when x=−2.

\(\ \begin{aligned} (x+1)^{(x-4)} &=1 \\ (-2+1)^{(-2-4)}-1 &=(-1)^{-6}-1 \\ &=\frac{1}{(-1)^{6}}-1 \\ &=0 \end{aligned}\)

The reason why this exercise is included is because you should not fall into the habit of assuming that you can take the log of both sides of an equation. It is only valid when the argument is strictly positive. For example, log(−2+1) (−2−4) =log(−1) is not possible.

Light intensity, measured in lumens, can be described by the relationship between i for intensity and d for depth in feet as it travels at specific depths of water in a swimming pool. What is the intensity of light at 10 feet?

\(\ \log \left(\frac{i}{12}\right)=-0.0145 \cdot d\)

Given d=10, solve for i measured in lumens.

\(\ \begin{aligned} \log \left(\frac{i}{12}\right) &=-0.0145 \cdot d \\ \log \left(\frac{i}{12}\right) &=-0.0145 \cdot 10 \\ \log \left(\frac{i}{12}\right) &=-0.145 \\ \left(\frac{i}{12}\right) &=10^{-0.145} \\ i &=12 \cdot 10^{-0.145} \approx 8.594 \end{aligned}\)

Solve the following equation for all possible values of x.

\(\ \frac{e^{x}-e^{-x}}{3}=14\)

First solve for e x ,

\(\ \begin{aligned} \frac{e^{x}-e^{-x}}{3} &=14 \\ e^{x}\left(e^{x}-e^{-x}\right) &=(42) e^{x} \\ e^{2 x}-1 &=42 e^{x} \\ \left(e^{x}\right)^{2}-42 e^{x}-1 &=0 \end{aligned}\)

Let u=e x .

\(\ \begin{array}{l} u^{2}-42 u-1=0 \\ u=\frac{-(-42) \pm \sqrt{(-42)^{2}-4 \cdot 1 \cdot(-1)}}{2 \cdot 1}=\frac{42 \pm \sqrt{1768}}{2} \approx 42.023796,-0.0237960 \end{array}\)

Note that the negative result is extraneous because ex must be greater than zero, so you only proceed in solving for x for one result.

\(\ \begin{aligned} e^{x} & \approx 42.023796 \\ x & \approx \ln 42.023796 \approx 3.738 \end{aligned}\)

Solve the following equation for all possible values of x: (log 2 x) 2 −log2x 7 =−12.

In calculus it is common to use a small substitution to simplify the problem and then substitute back later. In this case let u=log 2 x. Notice that this is a quadratic problem.

\(\ \begin{aligned} \left(\log _{2} x\right)^{2}-7 \log _{2} x+12 &=0 \\ u^{2}-7 u+12 &=0 \\ (u-3)(u-4) &=0 \\ u &=3,4 \end{aligned}\)

Now substitute back and solve for x in each case.

\(\ \begin{array}{l} \log _{2} x=3 \leftrightarrow 2^{3}=x=8 \\ \log _{2} x=4 \leftrightarrow 2^{4}=x=16 \end{array}\)

Solve each equation for x. Round each answer to three decimal places.

3. 12 4x =1020

4. 7 3x =2400

5. 2 x+1 −5=22

6. 5x+12 x =5x+7

7. 2 x+1 =2 2x+3

8. 3 x+3 =9 x+1

9. 2 x+4 =5 x

10. 13⋅8 0.2x =546

11. b x =c+a

12. 32 x =0.94−.12

Solve each log equation by using log properties and rewriting as an exponential equation.

13. log 3 x+log 3 5=2

14. 2logx=log8+log5−log10

15. log 9 x=\(\ \frac{3}{2}\)

Review (Answers)

To see the Review answers, open this PDF file and look for section 3.6.

6.7 Exponential and Logarithmic Models

Learning objectives.

In this section, you will:

• Model exponential growth and decay.
• Use Newton’s Law of Cooling.
• Use logistic-growth models.
• Choose an appropriate model for data.
• Express an exponential model in base e e .

We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling.

Modeling Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

where A 0 A 0 is equal to the value at time zero, e e is Euler’s constant, and k k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time , the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form y = A 0 e k t y = A 0 e k t where A 0 A 0 is the starting value, and e e is Euler’s constant. Now k k is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life , or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in Figure 2 and Figure 3 . It is important to remember that, although parts of each of the two graphs seem to lie on the x -axis, they are really a tiny distance above the x -axis.

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri , measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 × 10 13 . 4.01134972 × 10 13 . So, we could describe this number as having order of magnitude 10 13 . 10 13 .

Characteristics of the Exponential Function, y = A 0 e k t y = A 0 e k t

An exponential function with the form y = A 0 e k t y = A 0 e k t has the following characteristics:

• one-to-one function
• horizontal asymptote: y = 0 y = 0
• domain: ( – ∞ ,   ∞ ) ( – ∞ ,   ∞ )
• range: ( 0 , ∞ ) ( 0 , ∞ )
• x intercept: none
• y-intercept: ( 0 , A 0 ) ( 0 , A 0 )
• increasing if k > 0 k > 0 (see Figure 4 )
• decreasing if k < 0 k < 0 (see Figure 4 )

Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find A 0 A 0 we use the fact that A 0 A 0 is the amount at time zero, so A 0 = 10. A 0 = 10. To find k , k , use the fact that after one hour ( t = 1 ) ( t = 1 ) the population doubles from 10 10 to 20. 20. The formula is derived as follows

so k = ln ( 2 ) . k = ln ( 2 ) . Thus the equation we want to graph is y = 10 e ( ln 2 ) t = 10 ( e ln 2 ) t = 10 · 2 t . y = 10 e ( ln 2 ) t = 10 ( e ln 2 ) t = 10 · 2 t . The graph is shown in Figure 5 .

The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude 10 4 . 10 4 . The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude 10 7 , 10 7 , so we could say that the population has increased by three orders of magnitude in ten hours.

We now turn to exponential decay . One of the common terms associated with exponential decay, as stated above, is half-life , the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

We find that the half-life depends only on the constant k k and not on the starting quantity A 0 . A 0 .

The formula is derived as follows

Since t , t , the time, is positive, k k must, as expected, be negative. This gives us the half-life formula

Given the half-life, find the decay rate.

• Write A = A o e k t . A = A o e k t .
• Replace A A by 1 2 A 0 1 2 A 0 and replace t t by the given half-life.
• Solve to find k . k . Express k k as an exact value (do not round).

Note: It is also possible to find the decay rate using k = − ln ( 2 ) t . k = − ln ( 2 ) t .

Finding the Function that Describes Radioactive Decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t . t .

This formula is derived as follows.

The function that describes this continuous decay is f ( t ) = A 0 e ( ln ( 0.5 ) 5730 ) t . f ( t ) = A 0 e ( ln ( 0.5 ) 5730 ) t . We observe that the coefficient of t , t , ln ( 0.5 ) 5730 ≈ − 1.2097 × 10 −4 ln ( 0.5 ) 5730 ≈ − 1.2097 × 10 −4 is negative, as expected in the case of exponential decay.

The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time, measured in years.

Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating , which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t t years is

• A A is the amount of carbon-14 remaining
• A 0 A 0 is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

To find the age of an object, we solve this equation for t : t :

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r r be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation A ≈ A 0 e − 0.000121 t A ≈ A 0 e − 0.000121 t we know the ratio of the percentage of carbon-14 in the object we are dating to the initial amount of carbon-14 in the object when it was formed is r = A A 0 ≈ e − 0.000121 t . r = A A 0 ≈ e − 0.000121 t . We solve this equation for t , t , to get

Given the percentage of carbon-14 in an object, determine its age.

• Express the given percentage of carbon-14 as an equivalent decimal, k . k .
• Substitute for k in the equation t = ln ( r ) − 0.000121 t = ln ( r ) − 0.000121 and solve for the age, t . t .

Finding the Age of a Bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

We substitute 20 % = 0.20 20 % = 0.20 for r r in the equation and solve for t : t :

The bone fragment is about 13,301 years old.

The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as 13,301 years ± 1% or 13,301 years ± 133 years . 13,301 years ± 1% or 13,301 years ± 133 years .

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

Calculating Doubling Time

For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time .

Given the basic exponential growth equation A = A 0 e k t , A = A 0 e k t , doubling time can be found by solving for when the original quantity has doubled, that is, by solving 2 A 0 = A 0 e k t . 2 A 0 = A 0 e k t .

The formula is derived as follows:

Thus the doubling time is

Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

The function is A 0 e ln 2 2 t . A 0 e ln 2 2 t .

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

Using Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling , the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

• Newton’s Law of Cooling

The temperature of an object, T , T , in surrounding air with temperature T s T s will behave according to the formula

• t t is time
• A A is the difference between the initial temperature of the object and the surroundings
• k k is a constant, the continuous rate of cooling of the object

Given a set of conditions, apply Newton’s Law of Cooling.

• Set T s T s equal to the y -coordinate of the horizontal asymptote (usually the ambient temperature).
• Substitute the given values into the continuous growth formula T ( t ) = A e k t + T s T ( t ) = A e k t + T s to find the parameters A A and k . k .
• Substitute in the desired time to find the temperature or the desired temperature to find the time.

A cheesecake is taken out of the oven with an ideal internal temperature of 165°F, 165°F, and is placed into a 35°F 35°F refrigerator. After 10 minutes, the cheesecake has cooled to 150°F . 150°F . If we must wait until the cheesecake has cooled to 70°F 70°F before we eat it, how long will we have to wait?

Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation

We know the initial temperature was 165, so T ( 0 ) = 1 6 5 . T ( 0 ) = 1 6 5 .

We were given another data point, T ( 1 0 ) = 1 5 0 , T ( 1 0 ) = 1 5 0 , which we can use to solve for k . k .

This gives us the equation for the cooling of the cheesecake: T ( t ) = 1 3 0 e – 0 . 0 1 2 3 t + 3 5 . T ( t ) = 1 3 0 e – 0 . 0 1 2 3 t + 3 5 .

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70°F . 70°F .

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?

Using Logistic Growth Models

Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity . For constants a, b, a, b, and c, c, the logistic growth of a population over time t t is represented by the model

The graph in Figure 6 shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.

Logistic Growth

The logistic growth model is

• c 1 + a c 1 + a is the initial value
• c c is the carrying capacity , or limiting value
• b b is a constant determined by the rate of growth.

Using the Logistic-Growth Model

An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.

For example, at time t = 0 t = 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is b = 0.6030. b = 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

We substitute the given data into the logistic growth model

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is c = 1000. c = 1000. To find a , a , we use the formula that the number of cases at time t = 0 t = 0 is c 1 + a = 1 , c 1 + a = 1 , from which it follows that a = 999. a = 999. This model predicts that, after ten days, the number of people who have had the flu is f ( t ) = 1000 1 + 999 e − 0.6030 x ≈ 293.8. f ( t ) = 1000 1 + 999 e − 0.6030 x ≈ 293.8. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, c = 1000. c = 1000.

Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.

The graph in Figure 7 gives a good picture of how this model fits the data.

Using the model in Example 6 , estimate the number of cases of flu on day 15.

Choosing an Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed in Table 1 ? Find the model, and use a graph to check your choice.

First, plot the data on a graph as in Figure 8 . For the purpose of graphing, round the data to two decimal places.

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try y = a ln ( b x ) . y = a ln ( b x ) . Plugging in the first point, ( 1,0 ) , ( 1,0 ) , gives 0 = a ln b . 0 = a ln b . We reject the case that a = 0 a = 0 (if it were, all outputs would be 0), so we know ln ( b ) = 0. ln ( b ) = 0. Thus b = 1 b = 1 and y = a ln ( x ) . y = a ln ( x ) . Next we can use the point ( 9,4 .394 ) ( 9,4 .394 ) to solve for a : a :

Because a = 4.394 ln ( 9 ) ≈ 2 , a = 4.394 ln ( 9 ) ≈ 2 , an appropriate model for the data is y = 2 ln ( x ) . y = 2 ln ( x ) .

To check the accuracy of the model, we graph the function together with the given points as in Figure 9 .

We can conclude that the model is a good fit to the data.

Compare Figure 9 to the graph of y = ln ( x 2 ) y = ln ( x 2 ) shown in Figure 10 .

The graphs appear to be identical when x > 0. x > 0. A quick check confirms this conclusion: y = ln ( x 2 ) = 2 ln ( x ) y = ln ( x 2 ) = 2 ln ( x ) for x > 0. x > 0.

However, if x < 0 , x < 0 , the graph of y = ln ( x 2 ) y = ln ( x 2 ) includes a “extra” branch, as shown in Figure 11 . This occurs because, while y = 2 ln ( x ) y = 2 ln ( x ) cannot have negative values in the domain (as such values would force the argument to be negative), the function y = ln ( x 2 ) y = ln ( x 2 ) can have negative domain values.

Does a linear, exponential, or logarithmic model best fit the data in Table 2 ? Find the model.

Expressing an Exponential Model in Base e e

While powers and logarithms of any base can be used in modeling, the two most common bases are 10 10 and e . e . In science and mathematics, the base e e is often preferred. We can use laws of exponents and laws of logarithms to change any base to base e . e .

Given a model with the form y = a b x , y = a b x , change it to the form y = A 0 e k x . y = A 0 e k x .

• Rewrite y = a b x y = a b x as y = a e ln ( b x ) . y = a e ln ( b x ) .
• Use the power rule of logarithms to rewrite y as y = a e x ln ( b ) = a e ln ( b ) x . y = a e x ln ( b ) = a e ln ( b ) x .
• Note that a = A 0 a = A 0 and k = ln ( b ) k = ln ( b ) in the equation y = A 0 e k x . y = A 0 e k x .

Changing to base e

Change the function y = 2.5 ( 3.1 ) x y = 2.5 ( 3.1 ) x so that this same function is written in the form y = A 0 e k x . y = A 0 e k x .

Change the function y = 3 ( 0.5 ) x y = 3 ( 0.5 ) x to one having e e as the base.

Access these online resources for additional instruction and practice with exponential and logarithmic models.

• Logarithm Application – pH
• Exponential Model – Age Using Half-Life
• Exponential Growth Given Doubling Time
• Exponential Growth – Find Initial Amount Given Doubling Time

6.7 Section Exercises

With what kind of exponential model would half-life be associated? What role does half-life play in these models?

What is carbon dating? Why does it work? Give an example in which carbon dating would be useful.

With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?

Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of Cooling would be applied.

What is an order of magnitude? Why are orders of magnitude useful? Give an example to explain.

The temperature of an object in degrees Fahrenheit after t minutes is represented by the equation T ( t ) = 68 e − 0.0174 t + 72. T ( t ) = 68 e − 0.0174 t + 72. To the nearest degree, what is the temperature of the object after one and a half hours?

For the following exercises, use the logistic growth model f ( x ) = 150 1 + 8 e − 2 x . f ( x ) = 150 1 + 8 e − 2 x .

Find and interpret f ( 0 ) . f ( 0 ) . Round to the nearest tenth.

Find and interpret f ( 4 ) . f ( 4 ) . Round to the nearest tenth.

Find the carrying capacity.

Graph the model.

Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.

Rewrite f ( x ) = 1.68 ( 0.65 ) x f ( x ) = 1.68 ( 0.65 ) x as an exponential equation with base e e to five decimal places.

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic.

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in t t years is modeled by the equation P ( t ) = 1000 1 + 9 e − 0.6 t . P ( t ) = 1000 1 + 9 e − 0.6 t .

Graph the function.

What is the initial population of fish?

To the nearest tenth, what is the doubling time for the fish population?

To the nearest whole number, what will the fish population be after 2 2 years?

To the nearest tenth, how long will it take for the population to reach 900 ? 900 ?

What is the carrying capacity for the fish population? Justify your answer using the graph of P . P .

A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?

The formula for an increasing population is given by P ( t ) = P 0 e r t P ( t ) = P 0 e r t where P 0 P 0 is the initial population and r > 0. r > 0. Derive a general formula for the time t it takes for the population to increase by a factor of M .

Recall the formula for calculating the magnitude of an earthquake, M = 2 3 log ( S S 0 ) . M = 2 3 log ( S S 0 ) . Show each step for solving this equation algebraically for the seismic moment S . S .

What is the y -intercept of the logistic growth model y = c 1 + a e − r x ? y = c 1 + a e − r x ? Show the steps for calculation. What does this point tell us about the population?

Prove that b x = e x ln ( b ) b x = e x ln ( b ) for positive b ≠ 1. b ≠ 1.

Real-World Applications

For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.

To the nearest hour, what is the half-life of the drug?

Write an exponential model representing the amount of the drug remaining in the patient’s system after t t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram.

Using the model found in the previous exercise, find f ( 10 ) f ( 10 ) and interpret the result. Round to the nearest hundredth.

For the following exercises, use this scenario: A tumor is injected with 0.5 0.5 grams of Iodine-125, which has a decay rate of 1.15% 1.15% per day.

To the nearest day, how long will it take for half of the Iodine-125 to decay?

Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram.

A scientist begins with 250 250 grams of a radioactive substance. After 250 250 minutes, the sample has decayed to 32 32 grams. Rounding to five decimal places, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?

The half-life of Radium-226 is 1590 1590 years. What is the annual decay rate? Express the decimal result to four decimal places and the percentage to two decimal places.

The half-life of Erbium-165 is 10 .4 10 .4 hours. What is the hourly decay rate? Express the decimal result to four decimal places and the percentage to two decimal places.

A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 573 0 573 0 years.)

A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 1350 bacteria. Rounding to five decimal places, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 3 hours?

For the following exercises, use this scenario: A biologist recorded a count of 360 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes.

To the nearest whole number, what was the initial population in the culture?

Rounding to six decimal places, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?

For the following exercises, use this scenario: A pot of warm soup with an internal temperature of 100° 100° Fahrenheit was taken off the stove to cool in a 69° F 69° F room. After fifteen minutes, the internal temperature of the soup was 95° F . 95° F .

Use Newton’s Law of Cooling to write a formula that models this situation.

To the nearest minute, how long will it take the soup to cool to 80° F? 80° F?

To the nearest degree, what will the temperature be after 2 2 and a half hours?

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of 165°F 165°F and is allowed to cool in a 75°F 75°F room. After half an hour, the internal temperature of the turkey is 145°F . 145°F .

Write a formula that models this situation.

To the nearest degree, what will the temperature be after 50 minutes?

To the nearest minute, how long will it take the turkey to cool to 110° F? 110° F?

For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.

Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper: 10 − 10   W m 2 , 10 − 10   W m 2 , Vacuum: 10 − 4 W m 2 , 10 − 4 W m 2 , Jet: 10 2   W m 2 10 2   W m 2

Recall the formula for calculating the magnitude of an earthquake, M = 2 3 log ( S S 0 ) . M = 2 3 log ( S S 0 ) . One earthquake has magnitude 3 . 9 3 . 9 on the MMS scale. If a second earthquake has 75 0 75 0 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.

For the following exercises, use this scenario: The equation N ( t ) = 500 1 + 49 e − 0.7 t N ( t ) = 500 1 + 49 e − 0.7 t models the number of people in a town who have heard a rumor after t days.

How many people started the rumor?

To the nearest whole number, how many people will have heard the rumor after 3 days?

As t t increases without bound, what value does N ( t ) N ( t ) approach? Interpret your answer.

For the following exercise, choose the correct answer choice.

A doctor injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation?

• ⓐ f ( t ) = 13 ( 0.0805 ) t f ( t ) = 13 ( 0.0805 ) t
• ⓑ f ( t ) = 13 e 0.9195 t f ( t ) = 13 e 0.9195 t
• ⓒ f ( t ) = 13 e ( − 0.0839 t ) f ( t ) = 13 e ( − 0.0839 t )
• ⓓ f ( t ) = 4.75 1 + 13 e − 0.83925 t f ( t ) = 4.75 1 + 13 e − 0.83925 t

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• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
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Chapter 3: Exponential and Logarithmic Functions

Section 3.4: exponential and logarithmic equations, learning outcomes.

• Use like bases to solve exponential equations.
• Use logarithms to solve exponential equations.
• Use the definition of a logarithm to solve logarithmic equations.
• Use the one-to-one property of logarithms to solve logarithmic equations.
• Solve applied problems involving exponential and logarithmic equations.

Figure 1. Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr)

In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Use like bases to solve exponential equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b , S , and T , where [latex]b>0,\text{ }b\ne 1[/latex], [latex]{b}^{S}={b}^{T}[/latex] if and only if S  = T .

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation [latex]{3}^{4x - 7}=\frac{{3}^{2x}}{3}[/latex]. To solve for x , we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x :

A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations

For any algebraic expressions S  and T , and any positive real number [latex]b\ne 1[/latex],

How To: Given an exponential equation with the form [latex]{b}^{S}={b}^{T}[/latex], where S  and  T  are algebraic expressions with an unknown, solve for the unknown.

• Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[/latex].
• Use the one-to-one property to set the exponents equal.
• Solve the resulting equation, S  = T , for the unknown.

Example 1: Solving an Exponential Equation with a Common Base

Solve [latex]{2}^{x - 1}={2}^{2x - 4}[/latex].

[latex]\begin{align} {2}^{x - 1}&={2}^{2x - 4}&& \text{The common base is }2. \\ x - 1&=2x - 4&& \text{By the one-to-one property the exponents must be equal}. \\ x&=3 && \text{Solve for }x. \end{align}[/latex]

Solve [latex]{5}^{2x}={5}^{3x+2}[/latex].

Rewriting Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation [latex]256={4}^{x - 5}[/latex]. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for x :

How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it.

• Rewrite each side in the equation as a power with a common base.

Example 2: Solving Equations by Rewriting Them to Have a Common Base

Solve [latex]{8}^{x+2}={16}^{x+1}[/latex].

[latex]\begin{align}{8}^{x+2}&={16}^{x+1}\\ {\left({2}^{3}\right)}^{x+2}&={\left({2}^{4}\right)}^{x+1}&& \text{Write }8\text{ and }16\text{ as powers of }2.\\ {2}^{3x+6}&={2}^{4x+4}&& \text{To take a power of a power, multiply exponents}.\\ 3x+6&=4x+4&& \text{Use the one-to-one property to set the exponents equal}. \\ x&=2&& \text{Solve for }x.\end{align}[/latex]

Solve [latex]{5}^{2x}={25}^{3x+2}[/latex].

Example 3: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve [latex]{2}^{5x}=\sqrt{2}[/latex].

[latex]\begin{align}{2}^{5x}&={2}^{\frac{1}{2}}&& \text{Write the square root of 2 as a power of }2.\\ 5x&=\frac{1}{2}&& \text{Use the one-to-one property}. \\ x&=\frac{1}{10}&& \text{Solve for }x. \end{align}[/latex]

Solve [latex]{5}^{x}=\sqrt{5}[/latex].

[latex]x=\frac{1}{2}[/latex]

Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.

Example 4: Solving an Equation with Positive and Negative Powers

Solve [latex]{3}^{x+1}=-2[/latex].

This equation has no solution. There is no real value of x  that will make the equation a true statement because any power of a positive number is positive.

Analysis of the Solution

The figure below shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Solve [latex]{2}^{x}=-100[/latex].

The equation has no solution.

 Use logarithms to solve exponential equations

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since [latex]\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)[/latex] is equivalent to a  = b , we may apply logarithms with the same base on both sides of an exponential equation.

How To: Given an exponential equation in which a common base cannot be found, solve for the unknown.

• If one of the terms in the equation has base 10, use the common logarithm.
• If none of the terms in the equation has base 10, use the natural logarithm.
• Use the rules of logarithms to solve for the unknown.

Example 5: Solving an Equation Containing Powers of Different Bases

Solve [latex]{5}^{x+2}={4}^{x}[/latex].

[latex]\begin{align}{5}^{x+2}&={4}^{x}&& \text{There is no easy way to get the powers to have the same base}.\\ \mathrm{ln}{5}^{x+2}&=\mathrm{ln}{4}^{x}&& \text{Take ln of both sides}. \\ \left(x+2\right)\mathrm{ln}5&=x\mathrm{ln}4&& \text{Use laws of logs}.\\ x\mathrm{ln}5+2\mathrm{ln}5&=x\mathrm{ln}4&& \text{Use the distributive law}. \\ x\mathrm{ln}5-x\mathrm{ln}4&=-2\mathrm{ln}5&& \text{Get terms containing }x\text{ on one side, terms without }x\text{ on the other}. \\ x\left(\mathrm{ln}5-\mathrm{ln}4\right)&=-2\mathrm{ln}5&& \text{On the left hand side, factor out an }x.\\ x\mathrm{ln}\left(\frac{5}{4}\right)&=\mathrm{ln}\left(\frac{1}{25}\right)&& \text{Use the laws of logs}.\\ x&=\frac{\mathrm{ln}\left(\frac{1}{25}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}&& \text{Divide by the coefficient of }x. \end{align}[/latex]

You can also finish the equation from [latex]x\left(\mathrm{ln}5-\mathrm{ln}4\right)=-2\mathrm{ln}5[/latex] by dividing by the parentheses, giving an equivalent solution of:

[latex]\begin{align}x=\frac{-2\mathrm{ln}5}{\mathrm{ln}5-\mathrm{ln}4}\end{align}[/latex]

Solve [latex]{2}^{x}={3}^{x+1}[/latex].

[latex]x=\frac{\mathrm{ln}(3)}{\mathrm{ln}\left(\frac{2}{3}\right)}[/latex]

Is there any way to solve [latex]{2}^{x}={3}^{x}[/latex]?

Yes. The solution is x = 0.

Equations Containing [latex]e[/latex]

One common type of exponential equations are those with base e . This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e  on either side, we can use the natural logarithm to solve it.

How To: Given an equation of the form [latex]y=A{e}^{kt}[/latex], solve for  t .

• Divide both sides of the equation by A .
• Apply the natural logarithm of both sides of the equation.
• Divide both sides of the equation by k .

Example 6: Solve an Equation of the Form [latex]y=A{e}^{kt}[/latex]

Solve [latex]100=20{e}^{2t}[/latex].

[latex]\begin{align}100 & =20{e}^{2t}\\ 5 & ={e}^{2t}&& \text{Divide by the coefficient of the power.}\\ \mathrm{ln}5& =2t&& \text{Take ln of both sides. Use the fact that }\mathrm{ln}\left(x\right)\text{ and }{e}^{x}\text{ are inverse functions.}\\ t& =\frac{\mathrm{ln}5}{2}&& \text{Divide by the coefficient of }t\text{.}\end{align}[/latex]

Using laws of logs, we can also write this answer in the form [latex]t=\mathrm{ln}\sqrt{5}[/latex]. If we want a decimal approximation of the answer, we use a calculator.

Solve [latex]3{e}^{0.5t}=11[/latex].

[latex]t=2\mathrm{ln}\left(\frac{11}{3}\right)[/latex] or [latex]\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}[/latex]

Does every equation of the form [latex]y=A{e}^{kt}[/latex] have a solution?

No. There is a solution when [latex]k\ne 0[/latex], and when y and A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}[/latex].

Example 7: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[/latex]

Solve [latex]4{e}^{2x}+5=12[/latex].

[latex]\begin{align}4{e}^{2x}+5&=12\\ 4{e}^{2x}&=7&& \text{Combine like terms}.\\ {e}^{2x}&=\frac{7}{4}&& \text{Divide by the coefficient of the power}. \\ 2x&=\mathrm{ln}\left(\frac{7}{4}\right)&& \text{Take ln of both sides}. \\ x&=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)&& \text{Solve for }x.\end{align}[/latex]

Solve [latex]3+{e}^{2t}=7{e}^{2t}[/latex].

[latex]t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)[/latex]

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Example 8: Solving Exponential Functions in Quadratic Form

Solve [latex]{e}^{2x}-{e}^{x}=56[/latex].

[latex]\begin{align}{e}^{2x}-{e}^{x} & =56 \\ {e}^{2x}-{e}^{x}-56& =0&& \text{Get one side of the equation equal to zero}.\\ \left({e}^{x}+7\right)\left({e}^{x}-8\right) & =0 && \text{Factor by the FOIL method}. \\ {e}^{x}+7& =0{ or }{e}^{x}-8=0 && \text{If a product is zero, then one factor must be zero}.\\ {e}^{x}& =-7{\text{ or e}}^{x}=8&& \text{Isolate the exponentials}. \\ {e}^{x}& =8&& \text{Reject the equation in which the power equals a negative number}. \\ x& =\mathrm{ln}8&& \text{Solve the equation in which the power equals a positive number}. \end{align}[/latex]

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[/latex] because a positive number never equals a negative number. The solution [latex]x=\mathrm{ln}\left(-7\right)[/latex] is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Solve [latex]{e}^{2x}={e}^{x}+2[/latex].

[latex]x=\mathrm{ln}2[/latex]

Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

 Use the definition of a logarithm to solve logarithmic equations

We have already seen that every logarithmic equation [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation [latex]{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3[/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x :

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations

For any algebraic expression S and real numbers b and c , where [latex]b>0,\text{ }b\ne 1[/latex],

Example 9: Using Algebra to Solve a Logarithmic Equation

Solve [latex]2\mathrm{ln}x+3=7[/latex].

[latex]\begin{align}2\mathrm{ln}x+3&=7\\ 2\mathrm{ln}x&=4&& \text{Subtract 3}. \\ \mathrm{ln}x&=2&& \text{Divide by 2}.\\ x&={e}^{2}&& \text{Rewrite in exponential form}. \end{align}[/latex]

Solve [latex]6+\mathrm{ln}x=10[/latex].

[latex]x={e}^{4}[/latex]

Example 10: Solving an Equation Using The Product Property of Logarithms

Solve [latex]{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x+4\right)=5[/latex].

[latex]\begin{align}&{\mathrm{log}}_{2}\left(x(x+4\right)=5 && \text{Apply the product rule of logarithms}. \\ &{2}^{5}=x(x+4) && \text{Change to exponential form}.\\ &32={x}^{2}+4x && \text{Simplify}. \\ &{x}^2+4x-32=0 && \text{Set the equation equal to zero}. \\ &(x-4)(x+8)=0 && \text{Factor using FOIL}.\\&x - 4=0\text{ or }x+8=0 && \text{If a product is zero, one of the factors must be zero}. \\ &x=4\text{ or }x=-8 && \text{Solve for }x. \end{align}[/latex]

Check your solutions

There is only one solution: x  = 4. The solution x  = –8 is negative, but it does not check when substituted into the original equation because the argument of the logarithm functions is negative.

Solve [latex]{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(x+6\right)=3[/latex].

Example 11: Solving an Equation Using the Quotient Property of Logarithms

Solve [latex]\mathrm{log}\left(7x+6\right)=1+\mathrm{log}\left(x-1\right)[/latex].

[latex]\begin{align} &\mathrm{log}\left(7x+6\right)-\mathrm{log}\left(x-1\right)=1 && \text{Get both logarithms to one side}.\\&\mathrm{log}\dfrac{7x+6}{x-1}=1 && \text{Apply the quotient rule of logarithms}\\ &{10}^{1}=\dfrac{7x+6}{x-1} && \text{Change to exponential form}.\\ &10(x-1)=7x+6 && \text{Clear the fraction}. \\ &10x-10=7x+6 &&\text{Distribute}. \\ &x=\dfrac{16}{3} &&\text{Solve for }x. \end{align}[/latex]

There is only one solution: x  = [latex]\dfrac{16}{3}[/latex]. This solution makes the argument of the logarithm positive, so it is the only solution.

Solve [latex]\mathrm{log}\left(8x+5\right)=1+\mathrm{log}\left(x-7\right)[/latex].

x  = [latex]\dfrac{75}{2}[/latex]

Example 12: Using a Graph to Understand the Solution to a Logarithmic Equation

Solve [latex]\mathrm{ln}x=3[/latex].

[latex]\begin{align}\mathrm{ln}x&=3 \\ x&={e}^{3}&& \text{Use the definition of the natural logarithm.} \end{align}[/latex]

Figure 2 represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\approx 20[/latex]. A calculator gives a better approximation: [latex]{e}^{3}\approx 20.0855[/latex].

Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[/latex] to 2 decimal places.

[latex]x\approx 9.97[/latex]

 Use the one-to-one property of logarithms to solve logarithmic equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x  > 0, S  > 0, T  > 0 and any positive real number b , where [latex]b\ne 1[/latex],

For example,

So, if [latex]x - 1=8[/latex], then we can solve for x , and we get x  = 9. To check, we can substitute x  = 9 into the original equation: [latex]{\mathrm{log}}_{2}\left(9 - 1\right)={\mathrm{log}}_{2}\left(8\right)=3[/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation [latex]\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)[/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x :

To check the result, substitute x  = 10 into [latex]\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)[/latex].

A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

For any algebraic expressions S and T and any positive real number b , where [latex]b\ne 1[/latex],

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How To: Given an equation containing logarithms, solve it using the one-to-one property.

• Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form [latex]{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T[/latex].
• Use the one-to-one property to set the arguments equal.
• Solve the resulting equation, S =  T , for the unknown.

Example 13: Solving an Equation Using the One-to-One Property of Logarithms

Solve [latex]\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right)[/latex].

[latex]\begin{align}&\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right)\\ &{x}^{2}=2x+3&& \text{Use the one-to-one property of the logarithm}. \\ &{x}^{2}-2x - 3=0&& \text{Get zero on one side before factoring}.\\ &\left(x - 3\right)\left(x+1\right)=0&& \text{Factor using FOIL}. \\ &x - 3=0\text{ or }x+1=0&& \text{If a product is zero, one of the factors must be zero}. \\ &x=3\text{ or }x=-1 && \text{Solve for }x. \end{align}[/latex]

There are two solutions: x  = 3 or x  = –1. The solution x  = –1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Solve [latex]\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}1[/latex].

x  = 1 or x  = –1

Key Equations

Key concepts.

• We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
• When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown.
• When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown.
• When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side.
• We can solve exponential equations with base e , by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.
• After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.
• When given an equation of the form [latex]{\mathrm{log}}_{b}\left(S\right)=c[/latex], where S  is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[/latex], and solve for the unknown.
• We can also use graphing to solve equations with the form [latex]{\mathrm{log}}_{b}\left(S\right)=c[/latex]. We graph both equations [latex]y={\mathrm{log}}_{b}\left(S\right)[/latex] and y  = c on the same coordinate plane and identify the solution as the x- value of the intersecting point.
• When given an equation of the form [latex]{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T[/latex], where S  and T  are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation S  = T  for the unknown.
• Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.

Section 3.4 Homework Exercises

1. How can an exponential equation be solved?

2. When does an extraneous solution occur? How can an extraneous solution be recognized?

3. When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

For the following exercises, use like bases to solve the exponential equation.

4. [latex]{4}^{-3v - 2}={4}^{-v}[/latex]

5. [latex]64\cdot {4}^{3x}=16[/latex]

6. [latex]{3}^{2x+1}\cdot {3}^{x}=243[/latex]

7. [latex]{2}^{-3n}\cdot \frac{1}{4}={2}^{n+2}[/latex]

8. [latex]625\cdot {5}^{3x+3}=125[/latex]

9. [latex]\frac{{36}^{3b}}{{36}^{2b}}={216}^{2-b}[/latex]

10. [latex]{\left(\frac{1}{64}\right)}^{3n}\cdot 8={2}^{6}[/latex]

For the following exercises, use logarithms to solve.

11. [latex]{9}^{x - 10}=1[/latex]

12. [latex]2{e}^{6x}=13[/latex]

13. [latex]{e}^{r+10}-10=-42[/latex]

14. [latex]2\cdot {10}^{9a}=29[/latex]

15. [latex]-8\cdot {10}^{p+7}-7=-24[/latex]

16. [latex]7{e}^{3n - 5}+5=-89[/latex]

17. [latex]{e}^{-3k}+6=44[/latex]

18. [latex]-5{e}^{9x - 8}-8=-62[/latex]

19. [latex]-6{e}^{9x+8}+2=-74[/latex]

20. [latex]{2}^{x+1}={5}^{2x - 1}[/latex]

21. [latex]{e}^{2x}-{e}^{x}-132=0[/latex]

22. [latex]7{e}^{8x+8}-5=-95[/latex]

23. [latex]10{e}^{8x+3}+2=8[/latex]

24. [latex]4{e}^{3x+3}-7=53[/latex]

25. [latex]8{e}^{-5x - 2}-4=-90[/latex]

26. [latex]{3}^{2x+1}={7}^{x - 2}[/latex]

27. [latex]{e}^{2x}-{e}^{x}-6=0[/latex]

28. [latex]3{e}^{3 - 3x}+6=-31[/latex]

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

29. [latex]\mathrm{log}\left(\frac{1}{100}\right)=-2[/latex]

30. [latex]{\mathrm{log}}_{324}\left(18\right)=\frac{1}{2}[/latex]

For the following exercises, use the definition of a logarithm to solve the equation.

31. [latex]5{\mathrm{log}}_{7}n=10[/latex]

32. [latex]-8{\mathrm{log}}_{9}x=16[/latex]

33. [latex]4+{\mathrm{log}}_{2}\left(9k\right)=2[/latex]

34. [latex]2\mathrm{log}\left(8n+4\right)+6=10[/latex]

35. [latex]10 - 4\mathrm{ln}\left(9 - 8x\right)=6[/latex]

For the following exercises, use the product or quotient properties of logarithms to solve.

36. [latex]\mathrm{log}(x)+\mathrm{log}(x+9)=1[/latex]

37. [latex]\mathrm{log}(x)+\mathrm{log}(x+3)=1[/latex]

38. [latex]{\mathrm{log}}_{2}(x+5)=3-{\mathrm{log}}_{2}(x+7)[/latex]

39. [latex]{\mathrm{log}}_{3}(x+12)=3-{\mathrm{log}}_{3}(x+6)[/latex]

40. [latex]{\mathrm{log}}_{6}(x+26)-{\mathrm{log}}_{6}(x-9)=2[/latex]

41. [latex]{\mathrm{log}}_{4}(x+12)-{\mathrm{log}}_{4}(x-3)=2[/latex]

42. [latex]\mathrm{log}(4x+5)=1+\mathrm{log}(x-7)[/latex]

43. [latex]\mathrm{log}(8x+1)=1+\mathrm{log}(x-1)[/latex]

44. [latex]{\mathrm{log}}_{6}(x-7)+{\mathrm{log}}_{6}(x-2)-{\mathrm{log}}_{6}(x)=2[/latex]

45. [latex]{\mathrm{log}}_{2}(x-6)+{\mathrm{log}}_{2}(x-4)-{\mathrm{log}}_{2}(x)=2[/latex]

For the following exercises, use the one-to-one property of logarithms to solve.

46. [latex]\mathrm{ln}\left(10 - 3x\right)=\mathrm{ln}\left(-4x\right)[/latex]

47. [latex]{\mathrm{log}}_{13}\left(5n - 2\right)={\mathrm{log}}_{13}\left(8 - 5n\right)[/latex]

48. [latex]\mathrm{log}\left(x+3\right)-\mathrm{log}\left(x\right)=\mathrm{log}\left(74\right)[/latex]

49. [latex]\mathrm{ln}\left(-3x\right)=\mathrm{ln}\left({x}^{2}-6x\right)[/latex]

50. [latex]{\mathrm{log}}_{4}\left(6-m\right)={\mathrm{log}}_{4}3m[/latex]

51. [latex]\mathrm{ln}\left(x - 2\right)-\mathrm{ln}\left(x\right)=\mathrm{ln}\left(54\right)[/latex]

52. [latex]{\mathrm{log}}_{9}\left(2{n}^{2}-14n\right)={\mathrm{log}}_{9}\left(-45+{n}^{2}\right)[/latex]

53. [latex]\mathrm{ln}\left({x}^{2}-10\right)+\mathrm{ln}\left(9\right)=\mathrm{ln}\left(10\right)[/latex]

For the following exercises, solve each equation for x.

54. [latex]\mathrm{log}\left(x+12\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(12\right)[/latex]

55. [latex]\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x - 3\right)=\mathrm{ln}\left(7x\right)[/latex]

56. [latex]{\mathrm{log}}_{2}\left(7x+6\right)=3[/latex]

57. [latex]\mathrm{ln}\left(7\right)+\mathrm{ln}\left(2 - 4{x}^{2}\right)=\mathrm{ln}\left(14\right)[/latex]

58. [latex]{\mathrm{log}}_{8}\left(x+6\right)-{\mathrm{log}}_{8}\left(x\right)={\mathrm{log}}_{8}\left(58\right)[/latex]

59. [latex]\mathrm{ln}\left(3\right)-\mathrm{ln}\left(3 - 3x\right)=\mathrm{ln}\left(4\right)[/latex]

60. [latex]\mathrm{log_{3}}\left(3x\right)-\mathrm{log_{3}}\left(6\right)=\mathrm{log_{3}}\left(77\right)[/latex]

For the following exercises, solve the equation for x, if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

61. [latex]{\mathrm{log}}_{9}\left(x\right)-5=-4[/latex]

62. [latex]{\mathrm{log}}_{3}\left(x\right)+3=2[/latex]

63. [latex]\mathrm{ln}\left(3x\right)=2[/latex]

64. [latex]\mathrm{ln}\left(x - 5\right)=1[/latex]

65. [latex]\mathrm{log}\left(4\right)+\mathrm{log}\left(-5x\right)=2[/latex]

66. [latex]-7+{\mathrm{log}}_{3}\left(4-x\right)=-6[/latex]

67. [latex]\mathrm{ln}\left(4x - 10\right)-6=-5[/latex]

68. [latex]\mathrm{log}\left(4 - 2x\right)=\mathrm{log}\left(-4x\right)[/latex]

69. [latex]{\mathrm{log}}_{11}\left(-2{x}^{2}-7x\right)={\mathrm{log}}_{11}\left(x - 2\right)[/latex]

70. [latex]\mathrm{ln}\left(2x+9\right)=\mathrm{ln}\left(-5x\right)[/latex]

71. [latex]{\mathrm{log}}_{9}\left(3-x\right)={\mathrm{log}}_{9}\left(4x - 8\right)[/latex]

72. [latex]\mathrm{log}\left({x}^{2}+13\right)=\mathrm{log}\left(7x+3\right)[/latex]

73. [latex]\frac{3}{{\mathrm{log}}_{2}\left(10\right)}-\mathrm{log}\left(x - 9\right)=\mathrm{log}\left(44\right)[/latex]

74. [latex]\mathrm{ln}\left(x\right)-\mathrm{ln}\left(x+3\right)=\mathrm{ln}\left(6\right)[/latex]

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

75. An account with an initial deposit of $6,500 earns 7.25% annual interest, compounded continuously. How much will the account be worth after 20 years?

76. The formula for measuring sound intensity in decibels D is defined by the equation [latex]D=10\mathrm{log}\left(\frac{I}{{I}_{0}}\right)[/latex], where I is the intensity of the sound in watts per square meter and [latex]{I}_{0}={10}^{-12}[/latex] is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of [latex]8.3\cdot {10}^{2}[/latex] watts per square meter?

77. The population of a small town is modeled by the equation [latex]P=1650{e}^{0.5t}[/latex] where t is measured in years. In approximately how many years will the town’s population reach 20,000?

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate x to 3 decimal places.

78. [latex]1000{\left(1.03\right)}^{t}=5000[/latex] using the common log.

79. [latex]{e}^{5x}=17[/latex] using the natural log

80. [latex]3{\left(1.04\right)}^{3t}=8[/latex] using the common log

81. [latex]{3}^{4x - 5}=38[/latex] using the common log

82. [latex]50{e}^{-0.12t}=10[/latex] using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

83. [latex]7{e}^{3x - 5}+7.9=47[/latex]

84. [latex]\mathrm{ln}\left(3\right)+\mathrm{ln}\left(4.4x+6.8\right)=2[/latex]

85. [latex]\mathrm{log}\left(-0.7x - 9\right)=1+5\mathrm{log}\left(5\right)[/latex]

86. Atmospheric pressure P in pounds per square inch is represented by the formula [latex]P=14.7{e}^{-0.21x}[/latex], where x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile)

87. The magnitude M of an earthquake is represented by the equation [latex]M=\frac{2}{3}\mathrm{log}\left(\frac{E}{{E}_{0}}\right)[/latex] where E is the amount of energy released by the earthquake in joules and [latex]{E}_{0}={10}^{4.4}[/latex] is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing [latex]1.4\cdot {10}^{13}[/latex] joules of energy?

88. Use the definition of a logarithm along with the one-to-one property of logarithms to prove that [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex].

89. Recall the formula for continually compounding interest, [latex]y=A{e}^{kt}[/latex]. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm.

90. Recall the compound interest formula [latex]A=a{\left(1+\frac{r}{k}\right)}^{kt}[/latex]. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t.

91. Newton’s Law of Cooling states that the temperature T of an object at any time t can be described by the equation [latex]T={T}_{s}+\left({T}_{0}-{T}_{s}\right){e}^{-kt}[/latex], where [latex]{T}_{s}[/latex] is the temperature of the surrounding environment, [latex]{T}_{0}[/latex] is the initial temperature of the object, and k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm.

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

Wilson Honors Pre-Calculus and Trigonometry: Unit 3: Exponential and Logarithmic Functions

• Unit 1: Functions and their graphs
• Unit 4: Trigonometry
• Unit 4B Graphing Trig Functions
• Unit 5: Trig Identities
• Unit 6: Law of Sines and Cosines
• Unit 2: Polynomials
• Unit 3: Exponentials and Logarithms
• Unit 8 Matrices

10/18/17: Secion 3.1 graphing exponentials

p. 225 # 7, 13 – 16, 21, 22, 23, 24, 31, 32 35, 49, 50, 71a&c, 79

For 21, 22, 31, 32, 49, 50

Find asymptote, list transformations, and tell if function is increasing or decreasing.  

10/20/17: 3.2 Day 1 Logarithmic Expressions

p. 235 # 7-11, 16-19, 24, 25, 33-36, 58, 60, 72, 86, 92

• 3.2 day 1 notes

10/24/17: Section 3.2 Day 2 graphing logarithms

• 3.2 day 2 notes

p. 235 # 37, 39, 40, 42, 45-50, 73, 75, 78, 84, 91, 94

10/30/17: Section 3.3 properties of logarithms

p. 241 # 15-18, 24, 27, 28, 32, 35, 39, 46, 51, 54, 55, 56, 57, 63, 75, 80, 81, 82, 86, 87

11/1/17: Section 3.4 solving exponential and logarithm equations

p. 251 # 17, 19, 21, 24, 29, 37, 38, 43, 44, 49, 51, 59, 81, 88, 90, 93, 108

11/3/17: Section 3.5 exponential and logarithmic models

p. 262 # 15, 18, 23, 26, 29, 34, 37, 47, 49, 64, 69, 71

• Last Updated: Apr 30, 2019 1:44 PM
• URL: https://libguides.ops.org/WilsonHPC

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5.3.1: Graphs and Properties of Exponential Growth and Decay Functions (Exercises)

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• Page ID 39446

• Rupinder Sekhon and Roberta Bloom
• De Anza College

SECTION 5.3 PROBLEM SET: GRAPHS AND PROPERTIES OF EXPONENTIAL GROWTH AND DECAY FUNCTIONS

In questions 1-4, let \(t\) = time in years and \(y\) = the value at time \(t\) or \(y\) = the size of the population at time \(t\). The domain is the set of non-negative values for \(t\); \(t ≥ 0\), because \(y\) represents a physical quantity and negative values for time may not make sense. For each question:

• Write the formula for the function in the form \(y = ab^t\)
• Sketch the graph of the function and mark the coordinates of the y-intercept.

In questions 5-8, let \(t\) = time in years and \(y\) = the value at time \(t\) or \(y\) = the size of the population at time \(t\). The domain is the set of non-negative values for \(t\); \(t ≥ 0\), because \(y\) represents a physical quantity and negative values for time may not make sense. For each question:

• Write the formula for the function in the form \(y = ae^{kt}\)

For questions 9-12

• Sketch a graph of exponential function.
• List the coordinates of the y intercept.
• State the equation of any asymptotes and state the whether the function approaches the asymptote as x →∞ or as x→ −∞ .
• State the domain and range.