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# Solving Exponential Equations using Logarithms

We know we can solve algebraic equations by isolating the variable using inverse operations on each side. For example, solving $x+7=10$ entails subtracting 7 from each side. Since logarithms are the inverse of exponents, we can solve exponential equations by taking the logarithm of both sides. In this article, we'll explore how to do so and look at a few sample problems to get some practice. Let's get started!

## The procedure for solving exponential equations using logarithms

There are generally three steps to using logarithms to solve exponential equations:

1. Isolate the exponential equation

We can only take logarithms of exponential equations, so any other math stuff in the equation must be dealt with beforehand.

2. Take the logarithm of each side

This step is where we actually log both sides.

3. Solve for the variable

Most logs will give us an approximate answer, so solving for the variable usually involves a calculator and a ≈ symbol instead of an = sign. Alternatively, we may express our answer in terms of logs.

## An example of solving exponential equations using logarithms

Let's start with a relatively straightforward practice problem by solving for x: ${2}^{x}=12$ . The exponential expression is the entire equation, so we can skip step 1 above. Step 2 is taking the log of both sides, which looks like this:

$\mathrm{log}\left({2}^{x}\right)=\mathrm{log}\left(12\right)$

We can simplify that:

$x×\mathrm{log}\left(2\right)=\mathrm{log}\left(12\right)$

Next, we need to divide $\mathrm{log}\left(12\right)$ by $\mathrm{log}\left(2\right)$ to determine what x is. Time to break out the calculator!

$\frac{\mathrm{log}\left(12\right)}{\mathrm{log}\left(2\right)}\approx 3.585$

Some problems may ask us to round to a specific decimal place, so watch out for that. Let's try another one: $8\left({10}^{x}\right)=12$

We have an extra 8 that isn't part of the exponential equation this time, so we need to take care of that before doing any logs. Divide each side of the equation by 8:

${10}^{x}=\frac{12}{8}=\frac{3}{2}$

Next, take the log of each side:

$\mathrm{log}\left({10}^{x}\right)=\mathrm{log}\left(\frac{3}{2}\right)$

We can simplify this:

$x×\mathrm{log}\left(10\right)=\mathrm{log}\left(\frac{3}{2}\right)$

$x=\frac{\mathrm{log}\left(\frac{3}{2}\right)}{\mathrm{log}\left(10\right)}\approx 0.176$

Let's try one more: ${e}^{\left(5x\right)}=30$

We're working with e this time, which means we're taking the ln of each side instead of a log. Let's do that:

$\mathrm{ln}\left({e}^{\left(5x\right)}\right)=\mathrm{ln}\left(30\right)$

Since ln is simply log_e, we can simplify it just as we did above:

$5x×\mathrm{ln}\left(e\right)=\mathrm{ln}\left(30\right)$

$x=\frac{\mathrm{ln}\left(30\right)}{5}\approx 0.680$

Finished! The base of the logarithm we use doesn't matter as long as we apply the correct one consistently.

## Practice questions on solving exponential equations using logarithms

Solve for x: ${5}^{x}=250$

We only have an exponential expression, so we can skip directly to taking the logs of both sides:

${\mathrm{log}}_{5}\left(250\right)=x$

We can simplify this:

${\mathrm{log}}_{5}\left(125×2\right)=x={\mathrm{log}}_{5}\left(125\right)+{\mathrm{log}}_{5}\left(2\right)$

We know that ${\mathrm{log}}_{5}\left(125\right)=3$ , so we can substitute that into our final answer:

$x=3+{\mathrm{log}}_{5}\left(2\right)$

## Flashcards covering the Solving Exponential Equations using Logarithms

Algebra II Flashcards

## Varsity Tutors can help with solving exponential equations using logarithms

Using logs to solve exponential equations is an extension of the basic PEMDAS most math students have been doing for years. That said, thinking in terms of exponents and logarithms doesn't always come naturally. If you need help realizing that log5 (125) = 3 or how we simplified the logarithms above, an experienced math tutor could pinpoint and address your specific areas of opportunity. They could also design sessions around your preferred study style for maximal efficiency. Reach out to the Educational Directors at Varsity Tutors today for more information.

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