# How do you solve  [(5^(2x+1)) - 2] * [(5^(x+1)) + 5] = 0?

Aug 9, 2015

This equation has one solution $x = \frac{{\log}_{5} 2 - 1}{2}$

#### Explanation:

On left side there is a multiplication of 2 expressions, so it is zero, when any of those expressions is:

${5}^{2 x + 1} - 2 = 0$ or ${5}^{x + 1} + 5 = 0$

${5}^{2 x + 1} = 2$ or ${5}^{x + 1} = - 5$

The second equation has no solutions, because exponential functions do not have negative values.

To solve the first equation we have to change 2 to a power of 5.

${5}^{2 x + 1} = {5}^{{\log}_{5} 2}$

Now, when we have the same base on both sides we can skip it to get

$2 x + 1 = {\log}_{5} 2$
$2 x = {\log}_{5} 2 - 1$

$x = \frac{{\log}_{5} 2 - 1}{2}$