# How do you solve 2^(2x+5) - 7(2^x)=0 ?

## I know this question is to do with logs, but not sure how to answer the question.

Aug 16, 2016

$x = \log \frac{\frac{7}{32}}{\log} 2 = - 2.193$, nearly..

#### Explanation:

Rearranging,

$2$(2x+5)/2^x=2(2x+5-x)=2^x2^5=32( 2^)x7>

So, ${2}^{x} = \frac{7}{32}$.

Equating logarithms and solving for x,

$x = \log \frac{\frac{7}{32}}{\log} 2$