# How do you solve 5^(x-1) = 2^x?

Aug 5, 2015

I found: $x = \frac{\ln 5}{\ln 5 - \ln 2}$

#### Explanation:

I would take the natural log ($\ln$) on both sides:
$\ln {5}^{x - 1} = \ln {2}^{x}$
then use the fact that:
$\ln {x}^{a} = a \ln x$
to get:
$\left(x - 1\right) \ln 5 = x \ln 2$
$x \ln 5 - \ln 5 - x \ln 2 = 0$
$x \left(\ln 5 - \ln 2\right) = \ln 5$
so that:
$x = \frac{\ln 5}{\ln 5 - \ln 2}$
This is equal to $= 1.7564$ (if you can use the calculator).