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### Course: Algebra (all content) > Unit 11

Lesson 25: Solving exponential equations with logarithms (Algebra 2 level)- Solving exponential equations using logarithms: base-10
- Solving exponential equations using logarithms
- Solve exponential equations using logarithms: base-10 and base-e
- Solving exponential equations using logarithms: base-2
- Solve exponential equations using logarithms: base-2 and other bases

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# Solving exponential equations using logarithms

Learn how to solve any exponential equation of the form a⋅b^(cx)=d. For example, solve 6⋅10^(2x)=48.

The key to solving exponential equations lies in logarithms! Let's take a closer look by working through some examples.

## Solving exponential equations of the form $a\cdot {b}^{x}=d$

Let's solve $5\cdot {2}^{x}=240$ .

To solve for $x$ , we must first isolate the exponential part. To do this, divide both sides by $5$ as shown below. We do $5$ and the $2$ as this goes against the order of operations!

**not**multiply theNow, we can solve for $x$ by converting the equation to logarithmic form.

And just like that we have solved the equation! The $x={\mathrm{log}}_{2}(48)$ .

**exact**solution isSince $48$ is not a rational power of $2$ , we must use the change of base rule and our calculators to evaluate the logarithm. This is shown below.

The $x\approx 5.585$ .

**approximate**solution, rounded to the nearest thousandth, is### Check your understanding

## Solving exponential equations of the form $a\cdot {b}^{cx}=d$

Let's take a look at another example. Let's solve $6\cdot {10}^{2x}=48$

We start again by isolating the exponential part by dividing both sides by $6$ .

Next, we can bring down the exponent by converting to logarithmic form.

Finally, we can divide both sides by $2$ to solve for $x$ .

This is the $10$ .

**exact**answer. To approximate the answer to the nearest thousandth, we can type this directly into the calculator. Notice here that there is no need to change the base since it is already in base### Check your understanding

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