Lesson Explainer: Solving Exponential Equations Using Logarithms Mathematics

In this explainer, we will learn how to use logarithms to solve exponential equations.

Letโ€™s start by considering the exponential equation 2=8๏—. We can see that 2 is being raised to the power of ๐‘ฅ on the left side. In other words, a variable occurs in an exponent. This is a common feature of all exponential equations.

Definition: Exponential Equation

An exponential equation is an equation in which a variable is used in one exponent or more.

Before looking at how to solve 2=8๏— using logarithms, letโ€™s explore two alternative methods we might use to solve it. We should get the same solution using these methods as those we get when using logarithms.

The first alternative method is to begin by defining the functions ๐‘“(๐‘ฅ)=2๏— and ๐‘”(๐‘ฅ)=8. We could then graph ๐‘ฆ=๐‘“(๐‘ฅ) and ๐‘ฆ=๐‘”(๐‘ฅ) on the same coordinate plane and determine the point of intersection of the graphs as shown.

The ๐‘ฅ-coordinate of the point of intersection is 3, so with this method we get a solution of ๐‘ฅ=3.

The second alternative method is to start by using the fact that 8 is a power of 2 to rewrite the equation. We know that 2=2ร—2ร—2=8๏Šฉ, so we could substitute 2๏Šฉ for 8 to get 2=2๏—๏Šฉ. Then we could equate the exponents to again get a solution of ๐‘ฅ=3.

Now letโ€™s consider how to solve 2=8๏— by using the relationship between exponential and logarithmic functions.

Definition: Logarithmic Function

A logarithmic function is the inverse of an exponential function. If ๐‘ฆ=๐‘Ž๏—, then ๐‘ฅ=๐‘ฆlog๏Œบ.

Since 2=8๏—, or 8=2๏—, is in the form ๐‘ฆ=๐‘Ž๏—, we know that the value of ๐‘ฆ is 8 and that the value of ๐‘Ž is 2. Thus, we can write the equation ๐‘ฅ=8log๏Šจ. To simplify the right side, we can ask ourselves, โ€œA base of 2 raised to what power is equal to 8?โ€ The answer is 3, so we get ๐‘ฅ=3, the same answer that we arrived at with the previous two methods.

Some exponential equations are more complex, however, so when solving them using logarithms, we often need to use one or more of the rules of logarithms below.

Properties: Rules of Logarithms

Product Rule: ย  logloglog๏Œบ๏Œบ๏Œบ๐‘š๐‘›=๐‘š+๐‘›

Quotient Rule: ย  logloglog๏Œบ๏Œบ๏Œบ๏€ป๐‘š๐‘›๏‡=๐‘šโˆ’๐‘›

Power Rule: ย  loglog๏Œบ๏‡๏Œบ๐‘š=๐‘˜๐‘š

Notice that, in each of the rules, the bases of the logarithms are the same on both sides of the equation.

  • The product rule states that the log of the product of two numbers is the sum of the log of the first factor and the log of the second factor. We would use the product rule to find that logloglog๏Šฉ๏Šฉ๏Šฉ(9โ‹…27)=9+27=2+3=5.
  • The quotient rule states that the log of the quotient of two numbers is the difference of the log of the dividend and the log of the divisor. We would use the quotient rule to find that logloglog๏Šจ๏Šจ๏Šจ๏€ผ6416๏ˆ=64โˆ’16=6โˆ’4=2.
  • The power rule states that the log of a base raised to a power is the product of the power and the log of the base. We would use the power rule to find that loglog๏Šช๏Šฎ๏Šช๏€น4๏…=84=8(1)=8.

When solving exponential equations using logarithms, we often use a base of 10 or a base of ๐‘’ for the log because of the buttons on our calculators. However, the base does not matter. Recall that when the base is 10, by convention, there is no need to specify it, and when the base is ๐‘’, we are taking the natural log. It is important to note that if we use a base of 10 or a base of ๐‘’ when taking the log of a number, often the result will not be an integer. This is fine, though, because one of the advantages of solving an exponential equation using logarithms is that the equation does not have to have an integer solution. We can use a scientific calculator to approximate the solution. Letโ€™s look at how we would do this in the examples that follow.

Example 1: Solving Exponential Equations Using Logarithms

Solve 3=11๏— for ๐‘ฅ, giving your answer to three decimal places.

Answer

Letโ€™s begin by recalling that a logarithmic function is the inverse of an exponential function. If ๐‘ฆ=๐‘Ž๏—, then ๐‘ฅ=๐‘ฆlog๏Œบ. Since 3=11๏—, or 11=3๏—, is in the form ๐‘ฆ=๐‘Ž๏—, we know that the value of ๐‘ฆ is 11 and that the value of ๐‘Ž is 3. Thus, we can write the equation ๐‘ฅ=11.log๏Šฉ

Since 11 is not a power of 3, we need to use a scientific calculator to simplify the right-hand side. We must make sure to use the appropriate keystrokes, keeping in mind that the base is 3 and not 10. In doing so, we get ๐‘ฅ=2.182658โ€ฆ, The problem asks for the value of ๐‘ฅ to three decimal places, so we need to consider the digit in the ten thousandths place, which is 6. Because this digit is greater than or equal to 5, we should round the digit 2 in the thousands place up to get an answer of ๐‘ฅโ‰ˆ2.183.

Another way to solve the equation 3=11๏— is to take the log of both sides. Using a base of 10 gives us loglog(3)=11.๏—

Notice that the base is not specified in the equation. The power rule of logarithms then allows us to rewrite log(3)๏— as ๐‘ฅโ‹…3log so that the equation becomes ๐‘ฅโ‹…3=11.loglog

Next, we can divide both sides by log3 to get ๐‘ฅ=113loglog and then use the log button on a scientific calculator to get ๐‘ฅ=113โ‰ˆ1.041390.47712โ‰ˆ2.182658โ€ฆloglog.

Just as before, we would round the digit 2 in the thousandths place up to arrive at an answer of ๐‘ฅโ‰ˆ2.183.

Note

When using a scientific calculator to approximate the value of the expression loglog113, it is important that we write both log11 and log3 to a sufficient number of decimal places if we calculate them separately. If, for example, we only wrote them to three decimal places, we would have got ๐‘ฅ=113โ‰ˆ1.0410.477โ‰ˆ2.182389โ€ฆ.loglog

In this case, we would have kept the 2 in the thousandths place instead of rounding it up, and we would have got an incorrect answer of ๐‘ฅโ‰ˆ2.182. For this reason, it is best to enter loglog113 into our calculator as a single expression rather than calculating each individual log.

Next, we will work on solving a problem involving an exponential equation with a binomial exponent.

Example 2: Solving Exponential Equations with Binomial Exponents Using Logarithms

Find, to the nearest hundredth, the value of ๐‘ฅ for which 2=9๏—๏Šฐ๏Šฎ.

Answer

In order to solve the equation for ๐‘ฅ, we can start by taking the log of both sides. If we use a base of 10, we will not have to specify it and will get the equation loglog๏€น2๏…=9.๏—๏Šฐ๏Šฎ

Using the power rule of logarithms, we can then rewrite log๏€น2๏…๏—๏Šฐ๏Šฎ as (๐‘ฅ+8)(2)log so that the equation becomes (๐‘ฅ+8)(2)=9.loglog

Dividing both sides of the equation by log2 gives us ๐‘ฅ+8=92,loglog and after subtracting 8 from both sides we get ๐‘ฅ=92โˆ’8.loglog

Now we can use the log button on a scientific calculator to help approximate the value of ๐‘ฅ. When doing so, it is best to enter loglog92โˆ’8 into the calculator as a single expression rather than finding log9 and log2 separately before simplifying. This way, there is no danger of rounding errors. Doing this gives us ๐‘ฅ=92โˆ’8=โˆ’4.830074999โ€ฆ,loglog which gives us an answer of ๐‘ฅโ‰ˆโˆ’4.83 to the nearest hundredth.

Now letโ€™s look at how to solve an equation with two binomial exponents instead of one.

Example 3: Solving Exponential Equations with Binomial Exponents Using Logarithms

Use a calculator to find the value of ๐‘ฅ for which 3=8๏Šฑ๏Šช๏—๏Šฑ๏Šฉ๏—๏Šฐ๏Šช๏Ž–๏Šญ. Give your answer correct to two decimal places.

Answer

To start, letโ€™s take the log of both sides of the equation. Using a base of 10, we get loglog๏€น3๏…=๏€น8๏….๏Šฑ๏Šช๏—๏Šฑ๏Šฉ๏—๏Šฐ๏Šช๏Ž–๏Šญ

The power rule of logarithms then allows us to rewrite log๏€น3๏…๏Šฑ๏Šช๏—๏Šฑ๏Šฉ as (โ€“4๐‘ฅโ€“3)(3)log and log๏€น8๏…๏—๏Šฐ๏Šช๏Ž–๏Šญ as (๐‘ฅ+4.7)(8)log, which gives us the equation (โ€“4๐‘ฅโ€“3)(3)=(๐‘ฅ+4.7)(8).loglog

After distributing log3 on the left side of the equation and log8 on the right side, the equation becomes โˆ’4๐‘ฅโ‹…3โˆ’33=๐‘ฅโ‹…8+4.78.loglogloglog

Now, in order to isolate the variable, letโ€™s move the terms containing an ๐‘ฅ to one side of the equation and the terms not containing an ๐‘ฅ to the other side. First, we will add 4๐‘ฅโ‹…3log to both sides to get โˆ’33=๐‘ฅโ‹…8+4.78+4๐‘ฅโ‹…3.loglogloglog

Next, we will subtract 4.78log from both sides so that the equation becomes โˆ’33โˆ’4.78=๐‘ฅโ‹…8+4๐‘ฅโ‹…3.loglogloglog

It is now possible for us to factor out an ๐‘ฅ from the right side of the equation, giving us โˆ’33โˆ’4.78=๐‘ฅ(8+43),loglogloglog and after we divide both sides of the equation by the expression loglog8+43, we arrive at โˆ’33โˆ’4.788+43=๐‘ฅ.loglogloglog

Finally, we can use the log button on a scientific calculator to enter the expression for ๐‘ฅ. The calculator gives us ๐‘ฅ=โˆ’2.018756992โ€ฆ, which gives us an answer of ๐‘ฅโ‰ˆโˆ’2.02 to two decimal places.

In the example that follows, we must also move the variable terms to one side of the equation.

Example 4: Solving Exponential Equations Using Logarithms

Solve 2โ‹…3=5โ‹…4๏—๏— for ๐‘ฅ, giving your answer to three decimal places.

Answer

The first step in solving 2โ‹…3=5โ‹…4๏—๏— for ๐‘ฅ is moving the terms with an exponent of ๐‘ฅ to one side of the equation and the terms without an exponent of ๐‘ฅ to the other side. If we divide both sides of the equation by 2โ‹…4๏—, we get 2โ‹…32โ‹…4=5โ‹…42โ‹…4.๏—๏—๏—๏—

We can see that there is now a 2 in both the numerator and the denominator of the fraction on the left side of the equation and a 4๏— in both the numerator and the denominator of the fraction on the right side. These terms will cancel out, giving us 34=52.๏—๏—

Recall that if two constants ๐‘Ž and ๐‘ are both raised to the power of ๐‘ฅ, then ๐‘Ž๐‘=๏€ป๐‘Ž๐‘๏‡๏—๏—๏—; so we can replace 34๏—๏— with ๏€ผ34๏ˆ๏—, getting ๏€ผ34๏ˆ=52.๏—

Because a logarithmic function is the inverse of an exponential function, we know that if ๐‘ฆ=๐‘Ž๏—, then ๐‘ฅ=๐‘ฆlog๏Œบ. The equation ๏€ผ34๏ˆ=52๏—, or 52=๏€ผ34๏ˆ๏—, is in the form ๐‘ฆ=๐‘Ž๏—, which tells us that the value of ๐‘ฆ is 52 and that the value of ๐‘Ž is 34. Thus, we can write the equation ๐‘ฅ=๏€ผ52๏ˆ.log๏Žข๏Žฃ

Now we can use a scientific calculator to simplify the right side, keeping in mind that the base is 34 and not 10. In doing so, we get ๐‘ฅ=โˆ’3.185081โ€ฆ, which gives us an answer of ๐‘ฅโ‰ˆโˆ’3.185 to three decimal places.

We can also take the log of both sides of the equation ๏€ผ34๏ˆ=52๏— to solve it. If we use a base of 10, we get loglog๏€ผ34๏ˆ=๏€ผ52๏ˆ.๏—

Using the power rule of logarithms, we can rewrite log๏€ผ34๏ˆ๏— as ๐‘ฅโ‹…๏€ผ34๏ˆlog, giving us the equation ๐‘ฅโ‹…๏€ผ34๏ˆ=๏€ผ52๏ˆ.loglog

Now, we can divide both sides by log๏€ผ34๏ˆ to get ๐‘ฅ=๏€ป๏‡๏€ป๏‡,loglog๏Šซ๏Šจ๏Šฉ๏Šช and then use the log button on a scientific calculator to enter the expression for ๐‘ฅ, giving us ๐‘ฅ=โˆ’3.185081โ€ฆ.

Again, we would keep the digit 5 in the thousandths place to arrive at an answer of ๐‘ฅโ‰ˆโˆ’3.185.

Finally, letโ€™s look at an example in which we must use two different rules of logarithms.

Example 5: Solving Exponential Equations with Binomial Exponents Using Logarithms

Use a calculator to find the value of ๐‘ฅ for which 2ร—7=16ร—7๏—๏—๏Šฐ๏Šฏ. Give your answer correct to two decimal places.

Answer

Letโ€™s begin by taking the log of both sides of the equation. Using a base of 10 gives us loglog(2ร—7)=๏€น16ร—7๏….๏—๏—๏Šฐ๏Šฏ

The product rule of logarithms allows us to rewrite log(2ร—7)๏— as loglog(2)+7๏— and log๏€น16ร—7๏…๏—๏Šฐ๏Šฏ as loglog16+๏€น7๏…๏—๏Šฐ๏Šฏ, which results in the equation loglogloglog(2)+7=16+๏€น7๏….๏—๏—๏Šฐ๏Šฏ

We can now use the power rule of logarithms to rewrite log(2)๏— as ๐‘ฅโ‹…2log. We can also use it to rewrite log๏€น7๏…๏—๏Šฐ๏Šฏ as (๐‘ฅ+9)7log, giving us the equation ๐‘ฅโ‹…2+7=16+(๐‘ฅ+9)7.loglogloglog

After we then distribute ๐‘ฅ+9 on the right side, we get ๐‘ฅโ‹…2+7=16+๐‘ฅโ‹…2+97.logloglogloglog

Next, we must move the terms containing an ๐‘ฅ to one side of the equation and the terms not containing an ๐‘ฅ to the other side. Subtracting ๐‘ฅโ‹…7log from both sides gives us ๐‘ฅโ‹…2+7โˆ’๐‘ฅโ‹…7=16+97,logloglogloglog and then subtracting log7 from both sides gives us ๐‘ฅโ‹…2โˆ’๐‘ฅโ‹…7=16+97โˆ’7.logloglogloglog

The last two terms on the right side can now be combined to get the equation ๐‘ฅโ‹…2โˆ’๐‘ฅโ‹…7=16+87.loglogloglog

It is now possible for us to factor out an ๐‘ฅ from the left side of the equation, giving us ๐‘ฅ(2โˆ’7)=16+87,loglogloglog and after we divide both sides of the equation by the expression loglog2โ€“7, we arrive at ๐‘ฅ=16+872โ€“7.loglogloglog

Finally, we can use the log button on a scientific calculator to enter the expression for ๐‘ฅ. The calculator gives us ๐‘ฅ=โˆ’14.63953707โ€ฆ, which gives us an answer of ๐‘ฅโ‰ˆโˆ’14.64 to two decimal places.

Now letโ€™s finish by recapping some key points.

Key Points

  • An exponential equation is an equation in which a variable is used in one exponent or more.
  • A logarithmic function is the inverse of an exponential function. If ๐‘ฆ=๐‘Ž๏—, then ๐‘ฅ=๐‘ฆlog๏Œบ.
  • When solving exponential equations using logarithms, we must often use one or more of the rules of logarithms. Three logarithmic rules that are used when solving exponential equations are the product rule, the quotient rule, and the power rule.
  • The product rule, or logloglog๏Œบ๏Œบ๏Œบ๐‘š๐‘›=๐‘š+๐‘›, states that the log of the product of two numbers is the sum of the log of the first factor and the log of the second factor.
  • The quotient rule, or logloglog๏Œบ๏Œบ๏Œบ๏€ป๐‘š๐‘›๏‡=๐‘šโˆ’๐‘›, states that the log of the quotient of two numbers is the difference of the log of the dividend and the log of the divisor.
  • The power rule, or loglog๏Œบ๏‡๏Œบ๐‘š=๐‘˜๐‘š, states that the log of a base raised to a power is the product of the power and the log of the base.
  • It is best to enter expressions containing more than one log into our scientific calculator as a single expression rather than calculating each individual log. This helps prevent rounding errors.

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