Learning Objectives
In this section, you will:
- Use logarithms to solve exponential equations.
- Use the definition of a logarithm to solve logarithmic equations.
- Solve applied problems involving exponential and logarithmic equations.
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 equations.
Solving Exponential Equations Using Logarithms
Many times we need to solve equations of the form [latex]n=ab^x[/latex] where [latex]n[/latex] is a real number. To do this we will divide both sides by [latex]a[/latex] to get [latex]\frac{n}{a}=b^x[/latex] and then take the logarithm of both sides giving the equation [latex]\mathrm{log}\left(\frac{n}{a}\right)=\mathrm{log}\left(b^x\right).[/latex] Next, we use the power rule for logarithms to get [latex]\mathrm{log}\left(\frac{n}{a}\right)=x\mathrm{log}\left(b\right)[/latex]. Finally, divide both sides by [latex]\mathrm{log}\left(b\right)[/latex] to get [latex]x=\frac{\mathrm{log}\left(\frac{n}{a}\right)}{\mathrm{log}\left(b\right)}[/latex]. Note that any base for the logarithm can be used but base 10 and base [latex]e[/latex] are most commonly used.
How To
Given an exponential equation in the form [latex]n=ab^x[/latex], solve for the unknown.
- Divide both sides by [latex]a[/latex] or the initial condition.
- Apply the logarithm of both sides of the equation.
- 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 1: Solving a Basic Exponential Equation
Solve [latex]5=3\left(2\right)^x.[/latex]
Show Solution
Begin by dividing both sides by 3 to get
[latex]5/3=\left(2\right)^x.[/latex]
Then, take the natural logarithm of both sides to get the equation
[latex]\mathrm{ln}\left(5/3\right)=\mathrm{ln}\left(2^x\right).[/latex]
Use the power rule of logarithms [latex]\mathrm{ln}\left(b^x\right)=x\mathrm{ln}\left(b\right)[/latex] to get
[latex]\mathrm{ln}\left(5/3\right)=x\mathrm{ln}\left(2\right).[/latex]
Finally, divide both sides by [latex]\mathrm{ln}\left(2\right)[/latex] to get
[latex]x=\frac{\mathrm{ln}\left(5/3\right)}{\mathrm{ln}\left(2\right)}\approx0.7370.[/latex]
Try It #1
Solve [latex]7=15\left(4\right)^x.[/latex]
Show Solution
[latex]x=\frac{\mathrm{ln}\left(7/15\right)}{\mathrm{ln}\left(4\right)}\approx-0.5498.[/latex]
Example 2: Solving an Exponential Equation with an Algebraic Expression in the Exponent
Solve [latex]15=3\left(0.5\right)^{x+1}.[/latex]
Show Solution
Since the algebraic expression is in the exponent, this equation will be solved using the same steps as above but the x+1 should be kept in parenthesis until we are ready to simplify at the end. Begin by dividing both sides by 3 to get
[latex]5=\left(0.5\right)^{\left(x+1\right)}.[/latex]
Then, take the natural logarithm of both sides to get the equation
[latex]\mathrm{ln}\left(5\right)=\mathrm{ln}\left(0.5^{\left(x+1\right)}\right).[/latex]
Use the power rule of logarithms [latex]\mathrm{ln}\left(b^x\right)=x\mathrm{ln}\left(b\right)[/latex] so
[latex]\mathrm{ln}\left(5\right)={\left(x+1\right)}\mathrm{ln}\left(0.5\right).[/latex]
Divide both sides of the equation by [latex]\mathrm{ln}\left(0.5\right)[/latex] to get the equation
[latex]\left(x+1\right)=\frac{\mathrm{ln}\left(5\right)}{\mathrm{ln}\left(0.5\right)}.[/latex]
Note that this is just a linear equation so we subtract 1 from both sides and the solution is
[latex]x=\frac{\mathrm{ln}\left(5\right)}{\mathrm{ln}\left(0.5\right)}-1\approx-3.3219.[/latex]
Try It #2
Solve [latex]15=2\left(7\right)^{x^2+1}.[/latex]
Show Solution
[latex]x=\pm\sqrt{\frac{\mathrm{ln}\left(7.5\right)}{\mathrm{ln}\left(7\right)}-1}\approx\pm0.1883[/latex]
Equations Containing e
One common type of exponential equations uses base [latex]e.[/latex] 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 [latex]e[/latex] 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}\text{,}[/latex] solve for [latex]t.[/latex]
- Divide both sides of the equation by [latex]a.[/latex]
- Apply the natural logarithm of both sides of the equation.
- Divide both sides of the equation by [latex]k.[/latex]
Example 3: Solve an Equation with Continuous Growth
Solve [latex]100=20{e}^{2t}.[/latex]
Show Solution
For this problem, use the natural logarithm since the equation contains base [latex]e.[/latex]
[latex]\begin{align*}100&=20{e}^{2t}&& \hfill \\5&={e}^{2t}&&\textrm{Divide both sides by 20}\text{.}\hfill \\ \mathrm{ln}\left(5\right)&=2t&&\textrm{Take the natural logarithm of both sides.}\\ \text{ }&\text{ } &&\text{Use the fact that }\mathrm{ln}\left(e^{x}\right)=x\text{.}\hfill \\t&=\frac{\mathrm{ln}\left(5\right)}{2}&&\textrm{Divide by the coefficient of }t\text{.}\hfill \end{align*}[/latex]
Analysis
Using laws of logs, we can also write this answer in the form [latex]t=\mathrm{ln}\sqrt[\leftroot{1}\uproot{2} ]{5}.[/latex] If we want a decimal approximation of the answer, we use a calculator.
Try it #3
Solve [latex]3{e}^{0.5t}=11.[/latex]
Show Solution
[latex]t=2\mathrm{ln}\left(\frac{11}{3}\right)[/latex] or [latex]\mathrm{ln}\left({\left(\frac{11}{3}\right)}^{2}\right)[/latex]
Q&A
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]
Show Solution
This equation has no solution. There is no real value of [latex]x[/latex] that will make the equation a true statement because any power of a positive number is positive.
Figure 2 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.
Try it #4
Solve [latex]{2}^{x}=-100.[/latex]
Show Solution
The equation has no solution. We know that [latex]{2}^{x}[/latex] is always greater than 0. Therefore, there is no value for the input [latex]x[/latex] which produces an output of -100.
Sometimes there are exponential functions on both sides of the equation such as [latex]ab^x=cd^x[/latex]. The process is similar to solving [latex]n=ab^x[/latex] but we will need to be careful to use the product rule of logarithms before applying the power rule of logarithms.
How To
Given an equation with exponential expression on each side, solve for the unknown.
- Divide both sides by one of the initial conditions.
- Apply the logarithm of both sides of the equation.
- 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 product and power rules of logarithms and then solve for the unknown.
Example 5: Solving an Equation Containing Powers of Different Bases
Solve [latex]6\left(5\right)^{x+2}=2\left(4\right)^{x}.[/latex]
Show Solution
[latex]\begin{align*}6\left(5\right)^{x+2}&=2\left(4\right)^x&&\text{ }\\3\left(5\right)^{x+2}&=\left(4\right)^x&&\text{Divide both sides by 2.}\\\mathrm{ln}\left(3\left(5\right)^{x+2}\right)&=\mathrm{ln}\left(4^x\right)&&\text{Take the natural logarithm of both sides.}\\\mathrm{ln}\left(3\right)+\mathrm{ln}\left(5^{x+2}\right)&=\mathrm{ln}\left(4^x\right)&&\text{Use the product rule of logarithms.}\\\mathrm{ln}\left(3\right)+\left(x+2\right)\mathrm{ln}\left(5\right)&=x\mathrm{ln}\left(4\right)&&\text{Use the power law of logarithms.}\\\mathrm{ln}\left(3\right)+x\mathrm{ln}\left(5\right)+2\mathrm{ln}\left(5\right)&=x\mathrm{ln}\left(4\right)&&\text{Use the distributive law.}\\x\mathrm{ln}\left(5\right)-x\mathrm{ln}\left(4\right)&=-2\mathrm{ln}\left(5\right)-\mathrm{ln}\left(3\right)&&\text{Move terms containing x on one side, }\\&\text{}&&\text{and terms without x on the other.}\\x\left(\mathrm{ln}\left(5\right)-\mathrm{ln}\left(4\right)\right)&=-2\mathrm{ln}\left(5\right)-\mathrm{ln}\left(3\right)&& \text{On the left hand side, factor out an }x.\\x\mathrm{ln}\left(\frac{5}{4}\right)&=\mathrm{ln}\left(\frac{1}{75}\right)&&\text{Use the laws of logarithms.}\\x&=\frac{\mathrm{ln}\left(\frac{1}{75}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}&& \text{Divide by the coefficient of }x.\\\end{align*}[/latex]
Analysis
Notice that the product rule of logarithms was used before the power rule because the power rule cannot be applied to [latex]ab^x[/latex]; [latex]a[/latex] is not raised to the [latex]x[/latex] power. Also, note that when the power rule is used on [latex]\mathrm{ln}\left(5^{x+2}\right)[/latex] the [latex]\left(x+2\right)[/latex] has parenthesis so that the [latex]\mathrm{ln}\left(5\right)[/latex] gets properly distributed.
Try it #5
Solve [latex]{2}^{x}={3}^{x+1}.[/latex]
Show Solution
[latex]x=\frac{\mathrm{ln}\left(3\right)}{\mathrm{ln}\left(\frac{2}{3}\right)}[/latex]
Q&A
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 [latex]y[/latex] and [latex]A[/latex] 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 6: Solving an Equation That Requires Algebra First
Solve
- [latex]4{e}^{2x}+5=12.[/latex]
- [latex]{e}^{2t}-3=-4{e}^{2t}[/latex]
Show Solution
- [latex]\\[/latex][latex]\begin{align*}4{e}^{2x}+5&=12&&\text{ }\\4{e}^{2x}&=7&&\text{Combine like terms.}\\e^{2x}&=\frac{7}{4}&&\text{Divide by the coefficient.}\\2x&=\mathrm{ln}\left(\frac{7}{4}\right)&&\text{Take the natural logarithm of both sides.}\\x&=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)&&\text{Solve for }x. \end{align*}[/latex]
- [latex]\\[/latex][latex]\begin{align*}{e}^{2t}-3&=-4{e}^{2t}&&\text{ }\\5{e}^{2t}-3&=0&&\text{Add }4{e}^{2t}\text{ to both sides.}\\5{e}^{2t}&=3&&\text{Add 3 to both sides.}\\{e}^{2t}&=\frac{3}{5}&&\text{Divide both sides by 5.}\\2t&=\mathrm{ln}\left(\frac{3}{5}\right)&&\text{Take the natural logarithm of both sides.}\\t&=\frac{1}{2}\mathrm{ln}\left(\frac{3}{5}\right)&&\text{Solve for }x.\\\end{align*}[/latex]
Try it #6
Solve [latex]3+{e}^{2t}=7{e}^{2t}.[/latex]
Show Solution
[latex]t=\mathrm{ln}\left(\frac{1}{\sqrt[\leftroot{1}\uproot{2} ]{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.
These extraneous solutions frequently occur when the exponential function is a quadratic form. Recall that quadratic equations can be solved by factoring and setting each factor equal to zero, or by the quadratic equation. W will look for the pattern of the quadratic and then choose which technique can most easily be used.
Example 7: Solving Exponential Functions in Quadratic Form
Solve [latex]{e}^{2x}-{e}^{x}=56.[/latex]
Show Solution
We first re-write the equation using the fact that [latex]e^{2x}={\left({e^x}\right)}^2.[/latex] Notice that this equation has the form [latex]e^x[/latex] squared minus [latex]e^x[/latex] equals 56. This is a quadratic equation which can be solved by factoring.
[latex]\begin{align*}{e}^{2x}-{e}^{x}&=56&& \hfill \\ {e}^{2x}-{e}^{x}-56&=0&&\text{Get one side of the equation equal to zero}.\hfill \\ \left({e}^{x}+7\right)\left({e}^{x}-8\right)&=0&&\text{Factor by the FOIL method}.\hfill \\{e}^{x}+7&=0\text{ or }{e}^{x}-8=0&& \text{If a product is zero, then one factor must be zero}.\hfill \\{e}^{x}&=-7{\text{ or e}}^{x}=8&& \text{Isolate the exponentials}.\hfill \\ {e}^{x}&=8&&\text{Reject the equation that has no solution}.\hfill \\x&=\mathrm{ln}\left(8\right)&&\text{Write as a logarmithm}.\hfill \end{align*}[/latex]
Analysis
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]\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.
Try it #7
Solve [latex]{e}^{2x}={e}^{x}+2.[/latex]
Show Solution
[latex]x=\mathrm{ln}2.[/latex]
Using the Definition of a Logarithm to Solve Logarithmic Equations
We have already seen that the logarithmic equation [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex] is equivalent to the exponential equation [latex]{b}^{y}=x,[/latex] for [latex]x>0.[/latex] We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.
Example 8: Solving a Logarithmic Equation
Solve the equation [latex]{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x-5\right)=3.[/latex]
Show Solution
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 [latex]x:[/latex]
[latex]\begin{align*}\mathrm{log}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x-5\right)&=3&&\text{ }\\\mathrm{log}_{2}\left(2\left(3x-5\right)\right)&=3&&\text{Apply the product rule of logarithms}.\\{\mathrm{log}}_{2}\left(6x-10\right)&=3&&\text{Distribute}.\\{2}^{3}&=6x-10&&\text{Apply the definition of a logarithm}.\\8&=6x-10&&\text{Calculate }{2}^{3}.\\18&=6x&&\text{Add 10 to both sides}.\\x&=3&& \text{Divide by 6}.\end{align*}[/latex]
Example 9: Using Algebra to Solve a Logarithmic Equation
Solve [latex]2\mathrm{ln}\left(x\right)+3=7.[/latex]
Show Solution
[latex]\begin{align*}2\mathrm{ln}\left(x\right)+3&=7&&\text{ }\\2\mathrm{ln}\left(x\right)&=4&&\text{Subtract 3}.\\\mathrm{ln}\left(x\right)&=2&&\text{Divide by 2}.\\x&={e}^{2}&&\text{Rewrite in exponential form}.\end{align*}[/latex]
Try it #8
Solve [latex]6+\mathrm{ln}\left(x\right)=10.[/latex]
Example 10: Using Algebra Before and After Using the Definition of the Natural Logarithm
Solve [latex]2\mathrm{ln}\left(6x\right)=7.[/latex]
Show Solution
[latex]\begin{align*}2\mathrm{ln}\left(6x\right)&=7&&\text{ }\\\mathrm{ln}\left(6x\right)&=\frac{7}{2}&& \text{Divide by 2}.\\6x&={e}^{\left(\frac{7}{2}\right)}&&\text{Use the definition of the natural logarithm}.\\x&=\frac{1}{6}{e}^{\left(\frac{7}{2}\right)}&&\text{Divide by 6}.\end{align*}[/latex]
Try it #9
Solve [latex]2\mathrm{ln}\left(x+1\right)=10.[/latex]
Show Solution
[latex]x={e}^{5}-1[/latex]
Example 11: Using a Graph to Understand the Solution to a Logarithmic Equation
Solve [latex]\mathrm{ln}\left(x\right)=3.[/latex]
Show Solution
To solve [latex]\mathrm{ln}\left(x\right)=3,[/latex] write the equation in exponential form using the definition of the natural logarithm to get [latex]x={e}^{3}.[/latex][latex]\\[/latex]
Figure 3 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]
Try it #10
Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[/latex] to 2 decimal places.
Show Solution
[latex]x\approx 9.97[/latex]
Solving Applied Problems Using Exponential Equations
In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.
Example 12: Real World application; Depreciation
In 2018, a car was purchased for $32,000 and depreciates 20% per year. When will the car be worth $8,000?
Show Solution
We begin by writing a formula for the situation. Since a noncontinuous rate is given, we use the formula [latex]f\left(t\right)=ab^t.[/latex] Let [latex]t=0[/latex] represent 2019. The initial condition is [latex]a = 32000,[/latex] the rate is [latex]r=-0.2,[/latex] and the growth factor is [latex]b=1+r=1-0.2=0.8.[/latex] Therefore, the formula is
[latex]f\left(t\right)=32000\left(0.8\right)^t.[/latex]
Next, we need to know when the function’s output will be 8,000. Set the formula equal to 8000 and solve.
[latex]\begin{align*}8000&=32000\left(0.8\right)^t&&\text{ }\\0.25&=0.8^t&&\text{Divide both sides by 32000.}\\\mathrm{ln}\left(0.25\right)&=\mathrm{ln}\left(0.8^t\right)&&\text{Take the natural logarithm of both sides.}\\\mathrm{ln}\left(0.25\right)&=t\mathrm{ln}\left(0.8\right)&&\text{Use the power rule for logarithms.}\\t&=\frac{\mathrm{ln}\left(0.25\right)}{\mathrm{ln}\left(0.8\right)}\approx6.21&&\text{Divide both sides by }\mathrm{ln}\left(0.8\right).\end{align*}[/latex][latex]\\[/latex]
It will take approximately 6.2 years for the car to depreciate to $8,000. In 2024, the car will be worth $8,000.
Example 13: Real World Application; BLood Alcohol Content
A person’s blood alcohol content (BAC) is a measure of how much alcohol is in the bloodstream. When a person stops drinking, over time the BAC will decay exponentially. For a particular individual, the formula [latex]f\left(t\right)=0.1e^{-0.0067t}[/latex] models their BAC over time, t, measured in minutes. When will their BAC be 0.04?
Show Solution
We are asked to find the input when the output is 0.04 so solve the equation [latex]0.04=0.1e^{-0.0067t}.[/latex]
[latex]\begin{align*}0.4&=e^{-0.0067 t}&&\text{Divide both sides by 0.1.}\\\mathrm{ln}\left(0.4\right)&=-0.0067t&&\text{Take the natural logarithm and simplify.}\\t&=\frac{\mathrm{ln}\left(0.4\right)}{-0.0067}&&\text{Divide both sides by }-0.0067.\\\text{ }&\approx136.8&&\text{ }\end{align*}[/latex][latex]\\[/latex]
It will take approximately 137 minutes for the BAC to drop to 0.04 for this individual.
Example 14: Real World Application; Tuition
College A is charging $40,000 tuition in 2019 and it is increasing at a continuous rate of 7% per year. College B charges $45,000 in 2019, but it is increasing at 4% per year. When will college A cost more than college B?
Show Solution
First, we determine the model representing each of the tuition amounts. College A has a continuous rate so use [latex]A(t)=ae^{kt}[/latex] with [latex]a=40000[/latex] and [latex]k=0.07.[/latex] Therefore, the model for college A is [latex]A(t)=40000e^{0.07t}.[/latex] College B has a noncontinuous rate so use [latex]B(t)=ab^t[/latex] where [latex]a=45000[/latex] and [latex]b=1+r=1+0.04=1.04.[/latex] The model for college B is [latex]B(t)=45000\left(1.04\right)^t.[/latex]
We need to know when these two models are equal so we set the equations equal to each other and solve.
[latex]\begin{align*}40000e^{0.07t}&=45000\left(1.04\right)^t&&\text{ }\\\frac{8}{9}e^{0.07t}&=1.04^t&&\text{Divide both sides by 45000 and simplify.}\\\mathrm{ln}\left(\frac{8}{9}e^{0.07t}\right)&=\mathrm{ln}\left(1.04^t\right)&&\text{Take the natural logarithm of both sides.}\\\mathrm{ln}\left(\frac{8}{9}\right)+0.07t&=\mathrm{ln}\left(1.04^t\right)&&\text{Use the product rule and simplify.}\\\mathrm{ln}\left(\frac{8}{9}\right)+0.07t&=t\mathrm{ln}\left(1.04\right)&&\text{Use the power rule on the right side.}\\t\left(0.07-\mathrm{ln}\left(1.04\right)\right)&=-\mathrm{ln}\left(\frac{8}{9}\right)&&\text{Collect like terms and factor }t.\\t&=\frac{-\mathrm{ln}\left(\frac{8}{9}\right)}{0.07-\mathrm{ln}\left(1.04\right)}&&\text{Divide by the coefficient of }t.\\\text{ }&\approx3.83.&&\text{ }\\\end{align*}[/latex][latex]\\[/latex]
College B will have a higher tuition in approximately 4 years.
Try It #11
A town’s population is 14,000 people in 2017 and is increasing at a rate of 2.1% each year. When will the town’s population reach 18,000 people?
Show Solution
In approximately 12.1 years or in 2029, the population will be 18,000 people.
Conversions Between Continuous and Noncontinuous Growth Rates
Recall that there are two possible formulas that can be used to represent an exponential function: [latex]f(x)=ab^x=a{\left(b\right)}^x[/latex] for noncontinuous growth and [latex]f(x)=ae^{kx}=a{\left(e^k\right)}^x[/latex] for continuous growth. When comparing the two forms, we see that [latex]b=e^k.[/latex] Further, since [latex]b=1+r[/latex] where [latex]r[/latex] is the noncontinuous growth rate, we have that [latex]1+r=e^k.[/latex]
Example 15: Continuous Growth to Noncontinuous Growth
Given a continuous growth rate of 12%, find the noncontinuous rate.
Show Solution
The continuous growth rate [latex]k[/latex] is given as [latex]k=0.12.[/latex] We solve the equation [latex]1+r=e^k[/latex] with this value of [latex]k[/latex] plugged in.
[latex]\begin{align*}1+r&=e^{0.12}\\r&=e^{0.12}-1\approx0.1275.\end{align*}[/latex][latex]\\[/latex]
The noncontinuous rate is approximately 12.75%. Notice that since both rates are modeling the same growth, the noncontinuous rate must be slightly higher than the continuous rate, because the continuous rate allows for growth on the growth immediately.
Example 16: Noncontinuous Growth to Continuous Growth
Given a noncontinuous growth rate of 15%, find the continuous rate.
Show Solution
The noncontinuous growth rate [latex]r[/latex] is given as [latex]r=0.15.[/latex] Solve the equation [latex]1+r=e^k[/latex] with this value of [latex]r[/latex] plugged in.
[latex]\begin{align*}1+0.15&=e^{k}\\k&=\mathrm{ln}\left(1.15\right)\approx0.1398.\end{align*}[/latex][latex]\\[/latex]
We took the natural logarithm of both sides to solve the equation.
The continuous rate is approximately 13.98%.
Try It #12
a. Given a continuous decreasing rate of 5%, find the noncontinuous rate.
b. Given a noncontinuous increasing rate of 7%, find the continuous rate.
Show Solution
a. The noncontinuous rate of decrease is approximately 4.877% when the continuous rate of decrease is 5%.
b. The continuous rate of increase is approximately 6.766%, when the noncontinuous rate is 7%.
Solving Logarithmic Applications
Exponential growth and decay often involve very large or very small numbers. It is common to use a logarithmic scale when measurements result in extremely large values or extremely small values. The Richter Scale, pH and decibels are examples of such scales.
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 x 1013 . So, we could describe this number as having order of magnitude of 13.
The magnitude (size) of an earthquake is measured on a scale known as the Richter Scale. The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 6 is not twice as great as an earthquake of magnitude 3. It is 106−3 = 103 = 1,000 times as great!
The Richter scale strength of an earthquake, M, is given by [latex]M=\mathrm{log}\left(\frac{W}{W_0}\right),[/latex] where [latex]W[/latex] is the strength of the seismic waves of an earthquake and [latex]W_0[/latex] is the strength of normally occurring earthquakes. Minor earthquakes occur regularly allowing [latex]W_0[/latex] to be determined.
Example 17: The Richter Scale
An earthquake Richter Scale strength of 5 is considered a moderate strength earthquake. How much stronger was the 2010 magnitude 5.5 earthquake that occurred between Ontario and Quebec compared to a standard earthquake?
Show Solution
First we note that [latex]M=5.5[/latex] and substitute this into the Richter Scale equation. [latex]5.5=\mathrm{log}\left(\frac{W}{W_0}\right)[/latex] written as an exponential equation is [latex]10^{5.5}=\frac{W}{W_0}.[/latex] This can be written as [latex]W=10^{5.5}W_0[/latex] or [latex]W\approx316228W_0.[/latex]
The earthquake was 316,228 times stronger than a standard earthquake.
In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:
- Battery acid: 0.8
- Stomach acid: 2.7
- Orange juice: 3.3
- Pure water: 7 (at 25° C)
- Human blood: 7.35
- Fresh coconut: 7.8
- Sodium hydroxide (lye): 14
To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where [latex]\left[H^+\right][/latex] is the concentration of hydrogen ions in the solution measured in moles per liter.
[latex]pH=-\mathrm{log}\left(\left[H^+\right]\right)=\mathrm{log}\left(\frac{1}{\left[H^+\right]}\right)[/latex]
Example 18: pH
If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?
Show Solution
Suppose [latex]c[/latex] is the original concentration of hydrogen ions, and [latex]p[/latex] is the original pH of the liquid. Then [latex]p=-\mathrm{log}\left(c\right).[/latex] If the concentration is doubled, the new concentration is [latex]2c[/latex]. Then the pH of the new liquid is
[latex]pH=-\mathrm{log}\left(2c\right).[/latex]
Using the product rule of logarithms,
[latex]pH=-\left(\mathrm{log}\left(2\right)+\mathrm{log}\left(c\right)\right)=-\mathrm{log}\left(2\right)-\mathrm{log}\left(c\right).[/latex]
Since [latex]p=-log(c),[/latex] the new pH is
[latex]pH=p-\mathrm{log}\left(2\right)\approx p-0.301.[/latex]
When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.
Example 19: pH
Orange juice has a pH of approximately 3.3. Determine the hydrogen ion concentration.
Show Solution
Since the pH of orange juice is 3.3, we substitute 3.3 into the pH formula and get [latex]3.3=-\mathrm{log}\left(\left[H^+\right]\right).[/latex] Rewriting this as an exponential function, [latex]\left[H^+\right]=10^{-3.3}\approx5.012\times10^{-4}.[/latex] Therefore the concentration is [latex]\left[H^+\right]\approx5\times10^{-4}=0.0005,[/latex] or 0.0005 moles per liter.
A decibel (dB) is a measure of how loud a sound is when compared to a reference value. A commonly used reference value is the sound intensity of the softest sound a human can typically hear; usually that of a child. We will call this value [latex]I_0.[/latex] Since there is a very wide range of sounds that humans can hear, the logarithmic scale is used. The formula for decibels is
Sound level in decibels [latex]=10\mathrm{log}\left(\frac{I}{I_0}\right),[/latex]
where [latex]I[/latex] is the sound intensity of the sound being measured.
Example 20: Decibels
A vacuum cleaner sound level measures at 75dB and a balloon popping measures 125dB. A balloon popping is how many times more intense than the sound intensity of the vacuum cleaner?
Show Solution
Let [latex]I_B[/latex] be the sound intensity of the balloon popping and [latex]I_V[/latex] be the sound intensity of the vacuum cleaner. We then look at the difference.
[latex]\begin{align*}125\text{ dB }-75\text{ dB }&=10\mathrm{log}\left(\frac{I_B}{I_0}\right)-10\mathrm{log}\left(\frac{I_V}{I_0}\right)&&\text{ }\\50\text{ dB }&=10\mathrm{log}\left(\frac{\frac{I_B}{I_0}}{\frac{I_V}{I_0}}\right)&&\text{Use the quotient rule for logarithms.}\\5\text{ dB }&=\mathrm{log}\left(\frac{I_B}{I_V}\right)&&\text{Simplify.}\\10^5&=\frac{I_B}{I_V}&&\text{Write as an exponential equation.}\\100000I_V&=I_B&&\text{Multiply both sides by }I_V.\end{align*}[/latex][latex]\\[/latex]
The balloon’s sound intensity is 100,000 times more than the vacuum cleaner’s sound intensity.
Key Concepts
- An exponential equation can be solved by taking the logarithm of each side.
- We can solve exponential equations with base [latex]e,[/latex] 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 [latex]S[/latex] 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 [latex]y=c[/latex] on the same coordinate plane and identify the solution as the x-value of the intersecting point.
- 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.
Glossary
- extraneous solution
- a solution introduced while solving an equation that does not satisfy the conditions of the original equation
