# How do you solve 3^ { x } = \sqrt { 5^ { x - 2} }?

May 25, 2017

$x = {\log}_{\frac{5}{9}} \left(25\right)$

#### Explanation:

Let's rewrite the equation:
${3}^{x} = {\left({5}^{x - 2}\right)}^{\frac{1}{2}}$

${\left({3}^{x}\right)}^{2} = {\left({5}^{x - 2}\right)}^{\frac{1}{2} \cdot 2}$
${3}^{2 x} = \frac{{5}^{x}}{5} ^ 2$
$\frac{{5}^{x}}{3} ^ \left(2 x\right) = 25$
${\left(\frac{5}{9}\right)}^{x} = 25$

$x = {\log}_{\frac{5}{9}} \left(25\right) \approx - 5.48$

If you are unfamiliar with the concept of logarithm, here's a quick overview:

If we have ${b}^{x} = y$
Then, $x = {\log}_{b} \left(y\right)$, pronounced "log base b of y."