# How do you solve 8^(x-1)<=(1/2)^(2x-1) using a graph?

Feb 27, 2017

$x \le \frac{4}{5}$

#### Explanation:

First you reduce the expression to the same basis.

Knowing that $8 = {2}^{3}$ and $\frac{1}{2} = {2}^{-} 1$ we have

${2}^{3 \left(x - 1\right)} \le {2}^{- \left(2 x - 1\right)}$

${2}^{y}$ is a monotonically increasing function so this implies in

$3 \left(x - 1\right) \le - \left(2 x - 1\right)$ or

$5 x - 4 \le 0$ so $x \le \frac{4}{5}$ is the feasible set