solving exponential equations using logarithms common core algebra ii homework

Common Core Algebra II.Unit 4.Lesson 11.Solving Exponential Equations Using Logarithms

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Oct 7, 2016

Learning Common Core Algebra II.Unit 4.Lesson 11.Solving Exponential Equations Using Logarithms by eMathInstructions

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Solving Exponential Equations with Logs

How to solve exponential equations using logarithms.

In our previous lesson, you learned how to solve exponential equations without logarithms . This time around, we want to solve exponential equations requiring the use of logarithms . Why? The reason is that we can’t manipulate the exponential equation to have the same or common base on both sides of the equation. If you encounter such type of problem, the following are the suggested steps:

Steps to Solve Exponential Equations using Logarithms

1) Keep the exponential expression by itself on one side of the equation.

2) Get the logarithms of both sides of the equation. You can use any bases for logs.

3) Solve for the variable. Keep the answer exact or give decimal approximations. In addition to the steps above, make sure that you review the Basic Logarithm Rules because you will use them in one way or another.

Let’s go over some examples!

Examples of How to Solve Exponential Equations using Logarithms

Example 1: Solve the exponential equation [latex]{5^{2x}} = 21[/latex].

The good thing about this equation is that the exponential expression is already isolated on the left side. We can now take the logarithms of both sides of the equation. It doesn’t matter what base of the logarithm to use. The final answer should come out the same. The best choice for the base of log operation is [latex]5[/latex] since it is the base of the exponential expression itself. However, we will also use in the calculation the common base of [latex]10[/latex], and the natural base of [latex]\color{red}e[/latex] (denoted by [latex]\color{blue}ln[/latex]) just to show that in the end, they all have the same answers.

• Log Base of [latex]5[/latex]

• Log Base of [latex]10[/latex]

• Log Base of [latex]e[/latex]

Example 2: Solve the exponential equation [latex]2\left( {{3^{x – 5}}} \right) = 12[/latex] .

As you can see, the exponential expression on the left is not by itself. We must eliminate the number [latex]2[/latex] that is multiplying the exponential expression. To do that, divide both sides by [latex]2[/latex]. That would leave us just the exponential expression on the left, and [latex]6[/latex] on the right after simplification.

It’s time to take the log of both sides. Since the exponential expression has base [latex]3[/latex], that’s the convenient base to use for log operation. In addition, we will also solve this using the natural base [latex]e[/latex] just to compare if our final results agree.

• Log Base of [latex]3[/latex]

Example 3: Solve the exponential equation [latex]2\left({\Large{{{{{e^{4x – 3}}} \over {{e^{x – 2}}}}}}} \right) – 7 = 13[/latex] .

This looks like a mess at first. However, if you know how to start this out, the solution to this problem becomes a breeze. What we should do first is to simplify the expression inside the parenthesis. Use the Division Rule of Exponent by copying the common base of [latex]e[/latex] and subtracting the top by the bottom exponent.

Now isolate the exponential expression by adding both sides by [latex]7[/latex], followed by dividing the entire equation by [latex]2[/latex].

Take the logarithm of both sides. Use [latex]\color{red}ln[/latex] because we have a base of [latex]e[/latex]. Then solve for the variable [latex]x[/latex].

Example 4: Solve the exponential equation [latex]{1 \over 2}{\left( {{{10}^{x – 1}}} \right)^x} + 3 = 53[/latex] .

Observe that the exponential expression is being raised to [latex]x[/latex]. Simplify this by applying the Power to a Power Rule . Do that by copying the base [latex]10[/latex] and multiplying its exponent to the outer exponent. It should look like this after doing so.

We can now isolate the exponential expression by subtracting both sides by [latex]3[/latex] and then multiplying both sides by [latex]2[/latex].

Take the logarithm of both sides with base [latex]10[/latex]. If you just see a [latex]\color{red}log[/latex] without any specific base, it is understood to have [latex]10[/latex] as its base.

We are going to solve this quadratic equation by factoring method . Let’s move everything to the left side, therefore making the right side equal to zero. Factor out the trinomial into two binomials. Set each binomial factor equal zero then solve for [latex]x[/latex].

Example 5: Solve the exponential equation [latex]{e^{2x}} – 7{e^x} + 10 = 0[/latex].

We will need a different strategy to solve this exponential equation. Observe that we can actually convert this into a factorable trinomial. First, we let [latex]m = {e^x}[/latex]. Rewrite the exponential expression using this substitution.

Factor out the trinomial as a product of two binomials. Then replace [latex]m[/latex] by [latex]e^x[/latex] again.

Finally, set each factor equal to zero and solve for [latex]x[/latex], as usual, using logarithms.

You might also like these tutorials:

• Solving Exponential Equations without Logarithms

• Inspiration

Visit our websites: Math4kids CoolMathGames Teachers Parents Coding

Solving Exponential Equations

Now that you are getting the idea, what can we do to solve this one?

(I'm throwing a trick in, so be careful to clear the path!)

And, no, you cannot multiply the 2 and the 10 !

OK, now here's a tricky one!

Think!  You know how this works now...

WAIT!  The natural log can come to the rescue!

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10.5 Solve Exponential and Logarithmic Equations

Learning objectives.

• Solve logarithmic equations using the properties of logarithms
• Solve exponential equations using logarithms
• Use exponential models in applications

Be Prepared 10.5

Before you get started, take this readiness quiz.

• Solve: x 2 = 16 . x 2 = 16 . If you missed this problem, review Example 6.46 .
• Solve: x 2 − 5 x + 6 = 0 . x 2 − 5 x + 6 = 0 . If you missed this problem, review Example 6.45 .
• Solve: x ( x + 6 ) = 2 x + 5 . x ( x + 6 ) = 2 x + 5 . If you missed this problem, review Example 6.47 .

Solve Logarithmic Equations Using the Properties of Logarithms

In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations.

If our equation has two logarithms we can use a property that says that if log a M = log a N log a M = log a N then it is true that M = N . M = N . This is the One-to-One Property of Logarithmic Equations .

One-to-One Property of Logarithmic Equations

For M > 0 , N > 0 , a > 0 , M > 0 , N > 0 , a > 0 , and a ≠ 1 a ≠ 1 is any real number:

To use this property, we must be certain that both sides of the equation are written with the same base.

Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number.

Example 10.38

Solve: 2 log 5 x = log 5 81 . 2 log 5 x = log 5 81 .

2 log 5 x = log 5 81 Use the Power Property. log 5 x 2 = log 5 81 Use the One-to-One Property, if log a M = log a N , x 2 = 81 then M = N . Solve using the Square Root Property. x = ± 9 We eliminate x = −9 as we cannot take the logarithm x = 9 , x = −9 of a negative number. Check. x = 9 2 log 5 x = log 5 81 2 log 5 9 = ? log 5 81 log 5 9 2 = ? log 5 81 log 5 81 = log 5 81 ✓ 2 log 5 x = log 5 81 Use the Power Property. log 5 x 2 = log 5 81 Use the One-to-One Property, if log a M = log a N , x 2 = 81 then M = N . Solve using the Square Root Property. x = ± 9 We eliminate x = −9 as we cannot take the logarithm x = 9 , x = −9 of a negative number. Check. x = 9 2 log 5 x = log 5 81 2 log 5 9 = ? log 5 81 log 5 9 2 = ? log 5 81 log 5 81 = log 5 81 ✓

Try It 10.75

Solve: 2 log 3 x = log 3 36 2 log 3 x = log 3 36

Try It 10.76

Solve: 3 log x = log 64 3 log x = log 64

Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm.

Example 10.39

Solve: log 3 x + log 3 ( x − 8 ) = 2 . log 3 x + log 3 ( x − 8 ) = 2 .

log 3 x + log 3 ( x − 8 ) = 2 Use the Product Property, log a M + log a N = log a M ⋅ N . log 3 x ( x − 8 ) = 2 Rewrite in exponential form. 3 2 = x ( x − 8 ) Simplify. 9 = x 2 − 8 x Subtract 9 from each side. 0 = x 2 − 8 x − 9 Factor. 0 = ( x − 9 ) ( x + 1 ) Use the Zero-Product Property. x − 9 = 0 , x + 1 = 0 Solve each equation. x = 9 , x = −1 Check. x = −1 log 3 x + log 3 ( x − 8 ) = 2 log 3 ( −1 ) + log 3 ( −1 −8 ) = ? 2 We cannot take the log of a negative number. x = 9 log 3 x + log 3 ( x − 8 ) = 2 log 3 9 + log 3 ( 9 − 8 ) = ? 2 2 + 0 = ? 2 2 = 2 ✓ log 3 x + log 3 ( x − 8 ) = 2 Use the Product Property, log a M + log a N = log a M ⋅ N . log 3 x ( x − 8 ) = 2 Rewrite in exponential form. 3 2 = x ( x − 8 ) Simplify. 9 = x 2 − 8 x Subtract 9 from each side. 0 = x 2 − 8 x − 9 Factor. 0 = ( x − 9 ) ( x + 1 ) Use the Zero-Product Property. x − 9 = 0 , x + 1 = 0 Solve each equation. x = 9 , x = −1 Check. x = −1 log 3 x + log 3 ( x − 8 ) = 2 log 3 ( −1 ) + log 3 ( −1 −8 ) = ? 2 We cannot take the log of a negative number. x = 9 log 3 x + log 3 ( x − 8 ) = 2 log 3 9 + log 3 ( 9 − 8 ) = ? 2 2 + 0 = ? 2 2 = 2 ✓

Try It 10.77

Solve: log 2 x + log 2 ( x − 2 ) = 3 log 2 x + log 2 ( x − 2 ) = 3

Try It 10.78

Solve: log 2 x + log 2 ( x − 6 ) = 4 log 2 x + log 2 ( x − 6 ) = 4

When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed.

Example 10.40

Solve: log 4 ( x + 6 ) − log 4 ( 2 x + 5 ) = − log 4 x . log 4 ( x + 6 ) − log 4 ( 2 x + 5 ) = − log 4 x .

log 4 ( x + 6 ) − log 4 ( 2 x + 5 ) = − log 4 x Use the Quotient Property on the left side and the Power Property on the right. log 4 ( x + 6 2 x + 5 ) = log 4 x −1 Rewrite x −1 = 1 x . log 4 ( x + 6 2 x + 5 ) = log 4 1 x Use the One-to-One Property, if log a M = log a N , then M = N . x + 6 2 x + 5 = 1 x Solve the rational equation. x ( x + 6 ) = 2 x + 5 Distribute. x 2 + 6 x = 2 x + 5 Write in standard form. x 2 + 4 x − 5 = 0 Factor. ( x + 5 ) ( x − 1 ) = 0 Use the Zero-Product Property. x + 5 = 0 , x − 1 = 0 Solve each equation. x = −5 , x = 1 Check. We leave the check for you. log 4 ( x + 6 ) − log 4 ( 2 x + 5 ) = − log 4 x Use the Quotient Property on the left side and the Power Property on the right. log 4 ( x + 6 2 x + 5 ) = log 4 x −1 Rewrite x −1 = 1 x . log 4 ( x + 6 2 x + 5 ) = log 4 1 x Use the One-to-One Property, if log a M = log a N , then M = N . x + 6 2 x + 5 = 1 x Solve the rational equation. x ( x + 6 ) = 2 x + 5 Distribute. x 2 + 6 x = 2 x + 5 Write in standard form. x 2 + 4 x − 5 = 0 Factor. ( x + 5 ) ( x − 1 ) = 0 Use the Zero-Product Property. x + 5 = 0 , x − 1 = 0 Solve each equation. x = −5 , x = 1 Check. We leave the check for you.

Try It 10.79

Solve: log ( x + 2 ) − log ( 4 x + 3 ) = − log x . log ( x + 2 ) − log ( 4 x + 3 ) = − log x .

Try It 10.80

Solve: log ( x − 2 ) − log ( 4 x + 16 ) = log 1 x . log ( x − 2 ) − log ( 4 x + 16 ) = log 1 x .

Solve Exponential Equations Using Logarithms

In the section on exponential functions, we solved some equations by writing both sides of the equation with the same base. Next we wrote a new equation by setting the exponents equal.

It is not always possible or convenient to write the expressions with the same base. In that case we often take the common logarithm or natural logarithm of both sides once the exponential is isolated.

Example 10.41

Solve 5 x = 11 . 5 x = 11 . Find the exact answer and then approximate it to three decimal places.

5 x = 11 Since the exponential is isolated, take the logarithm of both sides. log 5 x = log 11 Use the Power Property to get the x as a factor, not an exponent. x log 5 = log 11 Solve for x . Find the exact answer. x = log 11 log 5 Approximate the answer. x ≈ 1.490 Since 5 1 = 5 and 5 2 = 25 , does it makes sense that 5 1.490 ≈ 11 ? 5 x = 11 Since the exponential is isolated, take the logarithm of both sides. log 5 x = log 11 Use the Power Property to get the x as a factor, not an exponent. x log 5 = log 11 Solve for x . Find the exact answer. x = log 11 log 5 Approximate the answer. x ≈ 1.490 Since 5 1 = 5 and 5 2 = 25 , does it makes sense that 5 1.490 ≈ 11 ?

Try It 10.81

Solve 7 x = 43 . 7 x = 43 . Find the exact answer and then approximate it to three decimal places.

Try It 10.82

Solve 8 x = 98 . 8 x = 98 . Find the exact answer and then approximate it to three decimal places.

When we take the logarithm of both sides we will get the same result whether we use the common or the natural logarithm (try using the natural log in the last example. Did you get the same result?) When the exponential has base e , we use the natural logarithm.

Example 10.42

Solve 3 e x + 2 = 24 . 3 e x + 2 = 24 . Find the exact answer and then approximate it to three decimal places.

3 e x + 2 = 24 Isolate the exponential by dividing both sides by 3. e x + 2 = 8 Take the natural logarithm of both sides. ln e x + 2 = ln 8 Use the Power Property to get the x as a factor, not an exponent. ( x + 2 ) ln e = ln 8 Use the property ln e = 1 to simplify. x + 2 = ln 8 Solve the equation. Find the exact answer. x = ln 8 − 2 Approximate the answer. x ≈ 0.079 3 e x + 2 = 24 Isolate the exponential by dividing both sides by 3. e x + 2 = 8 Take the natural logarithm of both sides. ln e x + 2 = ln 8 Use the Power Property to get the x as a factor, not an exponent. ( x + 2 ) ln e = ln 8 Use the property ln e = 1 to simplify. x + 2 = ln 8 Solve the equation. Find the exact answer. x = ln 8 − 2 Approximate the answer. x ≈ 0.079

Try It 10.83

Solve 2 e x − 2 = 18 . 2 e x − 2 = 18 . Find the exact answer and then approximate it to three decimal places.

Try It 10.84

Solve 5 e 2 x = 25 . 5 e 2 x = 25 . Find the exact answer and then approximate it to three decimal places.

Use Exponential Models in Applications

In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more options to solve these equations, we are able to solve more applications.

We will again use the Compound Interest Formulas and so we list them here for reference.

Compound Interest

For a principal, P , invested at an interest rate, r , for t years, the new balance, A is:

Example 10.43

Jermael’s parents put $10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth $50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?

A = $ 50,000 P = $ 10,000 Identify the variables in the formula. r = ? t = 17 years A = P e r t Substitute the values into the formula. 50,000 = 10,000 e r · 17 Solve for r . Divide each side by 10,000. 5 = e 17 r Take the natural log of each side. ln 5 = ln e 17 r Use the Power Property. ln 5 = 17 r ln e Simplify. ln 5 = 17 r Divide each side by 17. ln 5 17 = r Approximate the answer. r ≈ 0.095 Convert to a percentage. r ≈ 9.5 % They need the rate of growth to be approximately 9.5 % . A = $ 50,000 P = $ 10,000 Identify the variables in the formula. r = ? t = 17 years A = P e r t Substitute the values into the formula. 50,000 = 10,000 e r · 17 Solve for r . Divide each side by 10,000. 5 = e 17 r Take the natural log of each side. ln 5 = ln e 17 r Use the Power Property. ln 5 = 17 r ln e Simplify. ln 5 = 17 r Divide each side by 17. ln 5 17 = r Approximate the answer. r ≈ 0.095 Convert to a percentage. r ≈ 9.5 % They need the rate of growth to be approximately 9.5 % .

Try It 10.85

Hector invests $ 10,000 $ 10,000 at age 21. He hopes the investments will be worth $ 150,000 $ 150,000 when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal?

Try It 10.86

Rachel invests $ 15,000 $ 15,000 at age 25. She hopes the investments will be worth $ 90,000 $ 90,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula A = A 0 e k t . A = A 0 e k t . Exponential growth has a positive rate of growth or growth constant, k k , and exponential decay has a negative rate of growth or decay constant, k .

Exponential Growth and Decay

For an original amount, A 0 , A 0 , that grows or decays at a rate, k , for a certain time, t , the final amount, A , is:

We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.

Example 10.44

Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

This problem requires two main steps. First we must find the unknown rate, k . Then we use that value of k to help us find the unknown number of bacteria.

Identify the variables in the formula. A = 300 A 0 = 100 k = ? t = 3 hours A = A 0 e k t Substitute the values in the formula. 300 = 100 e k · 3 Solve for k . Divide each side by 100. 3 = e 3 k Take the natural log of each side. ln 3 = ln e 3 k Use the Power Property. ln 3 = 3 k ln e Simplify. ln 3 = 3 k Divide each side by 3. ln 3 3 = k Approximate the answer. k ≈ 0.366 We use this rate of growth to predict the number of bacteria there will be in 24 hours. A = ? A 0 = 100 k = ln 3 3 t = 24 hours A = A 0 e k t Substitute in the values. A = 100 e ln 3 3 · 24 Evaluate. A ≈ 656,100 At this rate of growth, they can expect 656,100 bacteria. Identify the variables in the formula. A = 300 A 0 = 100 k = ? t = 3 hours A = A 0 e k t Substitute the values in the formula. 300 = 100 e k · 3 Solve for k . Divide each side by 100. 3 = e 3 k Take the natural log of each side. ln 3 = ln e 3 k Use the Power Property. ln 3 = 3 k ln e Simplify. ln 3 = 3 k Divide each side by 3. ln 3 3 = k Approximate the answer. k ≈ 0.366 We use this rate of growth to predict the number of bacteria there will be in 24 hours. A = ? A 0 = 100 k = ln 3 3 t = 24 hours A = A 0 e k t Substitute in the values. A = 100 e ln 3 3 · 24 Evaluate. A ≈ 656,100 At this rate of growth, they can expect 656,100 bacteria.

Try It 10.87

Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

Try It 10.88

Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?

Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance.

Similar to the previous example, we can use the given information to determine the constant of decay, and then use that constant to answer other questions.

Example 10.45

The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years?

This problem requires two main steps. First we must find the decay constant k . If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find k . Then we use that value of k to help us find the amount of sample that will be left in 500 years.

Identify the variables in the formula. A = 50 A 0 = 100 k = ? t = 1590 years A = A 0 e k t Substitute the values in the formula. 50 = 100 e k · 1590 Solve for k . Divide each side by 100. 0.5 = e 1590 k Take the natural log of each side. ln 0.5 = ln e 1590 k Use the Power Property. ln 0.5 = 1590 k ln e Simplify. ln 0.5 = 1590 k Divide each side by 1590. ln 0.5 1590 = k exact answer We use this rate of growth to predict the amount that will be left in 500 years. A = ? A 0 = 100 k = ln 0.5 1590 t = 500 years A = A 0 e k t Substitute in the values. A = 100 e ln 0.5 1590 · 500 Evaluate. A ≈ 80.4 mg In 500 years there would be approximately 80.4 mg remaining. Identify the variables in the formula. A = 50 A 0 = 100 k = ? t = 1590 years A = A 0 e k t Substitute the values in the formula. 50 = 100 e k · 1590 Solve for k . Divide each side by 100. 0.5 = e 1590 k Take the natural log of each side. ln 0.5 = ln e 1590 k Use the Power Property. ln 0.5 = 1590 k ln e Simplify. ln 0.5 = 1590 k Divide each side by 1590. ln 0.5 1590 = k exact answer We use this rate of growth to predict the amount that will be left in 500 years. A = ? A 0 = 100 k = ln 0.5 1590 t = 500 years A = A 0 e k t Substitute in the values. A = 100 e ln 0.5 1590 · 500 Evaluate. A ≈ 80.4 mg In 500 years there would be approximately 80.4 mg remaining.

Try It 10.89

The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes?

Try It 10.90

The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?

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

• Solving Logarithmic Equations
• Solving Logarithm Equations
• Finding the rate or time in a word problem on exponential growth or decay

Section 10.5 Exercises

Practice makes perfect.

In the following exercises, solve for x .

log 4 64 = 2 log 4 x log 4 64 = 2 log 4 x

log 49 = 2 log x log 49 = 2 log x

3 log 3 x = log 3 27 3 log 3 x = log 3 27

3 log 6 x = log 6 64 3 log 6 x = log 6 64

log 5 ( 4 x − 2 ) = log 5 10 log 5 ( 4 x − 2 ) = log 5 10

log 3 ( x 2 + 3 ) = log 3 4 x log 3 ( x 2 + 3 ) = log 3 4 x

log 3 x + log 3 x = 2 log 3 x + log 3 x = 2

log 4 x + log 4 x = 3 log 4 x + log 4 x = 3

log 2 x + log 2 ( x − 3 ) = 2 log 2 x + log 2 ( x − 3 ) = 2

log 3 x + log 3 ( x + 6 ) = 3 log 3 x + log 3 ( x + 6 ) = 3

log x + log ( x + 3 ) = 1 log x + log ( x + 3 ) = 1

log x + log ( x − 15 ) = 2 log x + log ( x − 15 ) = 2

log ( x + 4 ) − log ( 5 x + 12 ) = − log x log ( x + 4 ) − log ( 5 x + 12 ) = − log x

log ( x − 1 ) − log ( x + 3 ) = log 1 x log ( x − 1 ) − log ( x + 3 ) = log 1 x

log 5 ( x + 3 ) + log 5 ( x − 6 ) = log 5 10 log 5 ( x + 3 ) + log 5 ( x − 6 ) = log 5 10

log 5 ( x + 1 ) + log 5 ( x − 5 ) = log 5 7 log 5 ( x + 1 ) + log 5 ( x − 5 ) = log 5 7

log 3 ( 2 x − 1 ) = log 3 ( x + 3 ) + log 3 3 log 3 ( 2 x − 1 ) = log 3 ( x + 3 ) + log 3 3

log ( 5 x + 1 ) = log ( x + 3 ) + log 2 log ( 5 x + 1 ) = log ( x + 3 ) + log 2

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

3 x = 89 3 x = 89

2 x = 74 2 x = 74

5 x = 110 5 x = 110

4 x = 112 4 x = 112

e x = 16 e x = 16

e x = 8 e x = 8

( 1 2 ) x = 6 ( 1 2 ) x = 6

( 1 3 ) x = 8 ( 1 3 ) x = 8

4 e x + 1 = 16 4 e x + 1 = 16

3 e x + 2 = 9 3 e x + 2 = 9

6 e 2 x = 24 6 e 2 x = 24

2 e 3 x = 32 2 e 3 x = 32

1 4 e x = 3 1 4 e x = 3

1 3 e x = 2 1 3 e x = 2

e x + 1 + 2 = 16 e x + 1 + 2 = 16

e x − 1 + 4 = 12 e x − 1 + 4 = 12

In the following exercises, solve each equation.

3 3 x + 1 = 81 3 3 x + 1 = 81

6 4 x − 17 = 216 6 4 x − 17 = 216

e x 2 e 14 = e 5 x e x 2 e 14 = e 5 x

e x 2 e x = e 20 e x 2 e x = e 20

log a 64 = 2 log a 64 = 2

log a 81 = 4 log a 81 = 4

ln x = −8 ln x = −8

ln x = 9 ln x = 9

log 5 ( 3 x − 8 ) = 2 log 5 ( 3 x − 8 ) = 2

log 4 ( 7 x + 15 ) = 3 log 4 ( 7 x + 15 ) = 3

ln e 5 x = 30 ln e 5 x = 30

ln e 6 x = 18 ln e 6 x = 18

3 log x = log 125 3 log x = log 125

7 log 3 x = log 3 128 7 log 3 x = log 3 128

log 6 x + log 6 ( x − 5 ) = log 6 24 log 6 x + log 6 ( x − 5 ) = log 6 24

log 9 x + log 9 ( x − 4 ) = log 9 12 log 9 x + log 9 ( x − 4 ) = log 9 12

log 2 ( x + 2 ) − log 2 ( 2 x + 9 ) = − log 2 x log 2 ( x + 2 ) − log 2 ( 2 x + 9 ) = − log 2 x

log 6 ( x + 1 ) − log 6 ( 4 x + 10 ) = log 6 1 x log 6 ( x + 1 ) − log 6 ( 4 x + 10 ) = log 6 1 x

In the following exercises, solve for x , giving an exact answer as well as an approximation to three decimal places.

6 x = 91 6 x = 91

( 1 2 ) x = 10 ( 1 2 ) x = 10

7 e x − 3 = 35 7 e x − 3 = 35

8 e x + 5 = 56 8 e x + 5 = 56

In the following exercises, solve.

Sung Lee invests $ 5,000 $ 5,000 at age 18. He hopes the investments will be worth $ 10,000 $ 10,000 when he turns 25. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation?

Alice invests $ 15,000 $ 15,000 at age 30 from the signing bonus of her new job. She hopes the investments will be worth $ 30,000 $ 30,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

Coralee invests $ 5,000 $ 5,000 in an account that compounds interest monthly and earns 7 % . 7 % . How long will it take for her money to double?

Simone invests $ 8,000 $ 8,000 in an account that compounds interest quarterly and earns 5 % . 5 % . How long will it take for his money to double?

Researchers recorded that a certain bacteria population declined from 100,000 to 100 in 24 hours. At this rate of decay, how many bacteria will there be in 16 hours?

Researchers recorded that a certain bacteria population declined from 800,000 to 500,000 in 6 hours after the administration of medication. At this rate of decay, how many bacteria will there be in 24 hours?

A virus takes 6 days to double its original population ( A = 2 A 0 ) . ( A = 2 A 0 ) . How long will it take to triple its population?

A bacteria doubles its original population in 24 hours ( A = 2 A 0 ) . ( A = 2 A 0 ) . How big will its population be in 72 hours?

Carbon-14 is used for archeological carbon dating. Its half-life is 5,730 years. How much of a 100-gram sample of Carbon-14 will be left in 1000 years?

Radioactive technetium-99m is often used in diagnostic medicine as it has a relatively short half-life but lasts long enough to get the needed testing done on the patient. If its half-life is 6 hours, how much of the radioactive material form a 0.5 ml injection will be in the body in 24 hours?

Writing Exercises

Explain the method you would use to solve these equations: 3 x + 1 = 81 , 3 x + 1 = 81 , 3 x + 1 = 75 . 3 x + 1 = 75 . Does your method require logarithms for both equations? Why or why not?

What is the difference between the equation for exponential growth versus the equation for exponential decay?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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Access for free at https://openstax.org/books/intermediate-algebra/pages/1-introduction

• Authors: Lynn Marecek
• Publisher/website: OpenStax
• Book title: Intermediate Algebra
• Publication date: Mar 14, 2017
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra/pages/10-5-solve-exponential-and-logarithmic-equations

© Feb 9, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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Unit 10 – Exponential and Logarithmic Functions

This unit is rich in theory and application.  Basic exponential functions are reviewed with the method of common bases introduced as their primary algebraic tool.  Exponential modeling of increasing and decreasing phenomena are extensively explored in two lessons.  Logarithms are introduced as the inverses of exponential functions.  Special care is taken to develop good number sense concerning logarithms before standard work is done with logarithm laws.  Equation work with logarithms emphasizes both solving equations that involve logarithms as well as solving exponential equations with logarithms.  The number e and the natural log are briefly introduced with the unit ending by revisiting regression in its exponential and logarithmic forms.

Introduction to Exponential Functions

LESSON/HOMEWORK

EDITABLE LESSON

EDITABLE KEY

The Method of Common Bases

Exponential Modeling – Day 1

Exponential Modeling – Day 2

Introduction to Logarithms

Graphs of Basic Logarithms

Logarithm Laws

Solving Equations Involving Logarithms

Solving Exponential Equations Using Logarithms

The Number e and the Natural Logarithm

Exponential and Logarithmic Regression

Unit Review

Unit 10 Review – Exponential and Logarithmic Functions

UNIT REVIEW

EDITABLE REVIEW

Unit 10 Assessment Form A

EDITABLE ASSESSMENT

Unit 10 Assessment Form B

Unit 10 Assessment Form C

Unit 10 Assessment Form D

Unit 10 Exit Tickets

Unit 10 – Mid-Unit Quiz (Through Lesson #6) – Form A

Unit 10 – Mid-Unit Quiz (Through Lesson #6) – Form B

Unit 10 – Mid-Unit Quiz (Through Lesson #6) – Form C

U10.AO.01 – Lesson 4.5 – Exponential Modeling Revisited

EDITABLE RESOURCE

U10.AO.02 – Lesson 7.5 – Log Law Practice

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