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

Sep 29, 2015

I found $x = 1.7564$ but....

#### Explanation:

I would try by....cheating a bit! I imagine that you can use a pocket calculator to evaluate the natural logarithm, $\ln$. So, I take the natural log of both sides:
$\ln {\left(5\right)}^{x - 1} = \ln {\left(2\right)}^{x}$
I use the rule of logs that tells us that: $\log {a}^{x} = x \log a$ to get:
$\left(x - 1\right) \ln \left(5\right) = x \ln 2$
$x \ln 5 - \ln 5 = x \ln 2$ rearranging:
$x \left(\ln 5 - \ln 2\right) = \ln 5$
$x \ln \left(\frac{5}{2}\right) = \ln \left(5\right)$
and $x = \ln \frac{5}{\ln \left(\frac{5}{2}\right)} =$ here comes the "cheating" bit:
$x = \frac{1.609}{0.916} = 1.7564$