# How do you solve 5^x= 20?

Dec 6, 2015

Take logs and use properties of logs to find:

$x = \frac{\log 20}{\log 5} = \frac{1 + \log 2}{1 - \log 2} \approx 1.86135$

#### Explanation:

If we take common logs of both sides then we get:

$\log 20 = \log {5}^{x} = x \log 5$

So $x = \frac{\log 20}{\log 5}$

We can do a little more with this if we know the log value $\log 2 \approx 0.30103$ and that $\log 10 = 1$:

$\log 5 = \log \left(\frac{10}{2}\right) = \log 10 - \log 2 = 1 - \log 2$

$\approx 1 - 0.30103 = 0.69897$

$\log 20 = \log \left(10 \cdot 2\right) = \log 10 + \log 2 = 1 + \log 2$

$\approx 1 + 0.30103 = 1.30103$

So $x = \frac{\log 20}{\log 5} \approx \frac{1.30103}{0.69897} \approx 1.86135$