Solving exponential equations

Sometimes the unknown is stuck up in the exponent, like this:

2^x = 20

We can't isolate x by ordinary rearranging — it isn't multiplied or added, it's an exponent. The trick is to take the logarithm of both sides, because a log turns a power into a product.

Taking \log of both sides keeps the equation balanced:

\log(2^x) = \log 20

Now use the power law, \log(a^x) = x\log a, to bring the exponent down in front:

x\log 2 = \log 20

With x back on the ground floor, it's an ordinary linear equation — just divide both sides by \log 2:

x = \frac{\log 20}{\log 2} \approx 4.32

That's the whole recipe: take logs, drop the power, divide. The same steps solve any b^x = k.

See it solved

Step through the solution to 2^x = 20: take logs of both sides, use the power law to bring the exponent down, then divide.

See it explained

Sal Khan solves an exponential equation by taking logarithms of both sides.