# How do you solve 5^ { 2x } + 20( 5^ { x } ) - 125= 0?

Mar 3, 2017

$x = 1$.

#### Explanation:

The equation can be written as

${\left({5}^{x}\right)}^{2} + 20 \left({5}^{x}\right) - 125 = 0$

We now let $u = {5}^{x}$.

${u}^{2} + 20 u - 125 = 0$

$\left(u + 25\right) \left(u - 5\right) = 0$

$u = - 25 \mathmr{and} 5$

${5}^{x} = - 25 \mathmr{and} {5}^{x} = 5$

There is no solution to the first solution (because if you take the $\ln$ of $- 25$, it is undefined). The second equation obviously has $x = 1$ as a solution.

Hopefully this helps!