# How do you solve for x in 2^x = 5^(x - 2)?

Jul 5, 2015

I found: $x = \frac{2 \ln \left(5\right)}{\ln \left(5\right) - \ln \left(2\right)}$

#### Explanation:

I would apply the natural logarithm, $\ln$, to both sides:
$\ln \left({2}^{x}\right) = \ln \left({5}^{x - 2}\right)$
apply one of the rules of logs:
$x \ln \left(2\right) = \left(x - 2\right) \ln \left(5\right)$
$x \ln \left(2\right) = x \ln \left(5\right) - 2 \ln \left(5\right)$ rearranging:
$x \left[\ln \left(5\right) - \ln \left(2\right)\right] = 2 \ln \left(5\right)$
and finally:
$x = \frac{2 \ln \left(5\right)}{\ln \left(5\right) - \ln \left(2\right)}$

Using a pocket calculator you can get $x = 3.513$