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

Question

2751 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T18:04:03

$x le 4/5$

First you reduce the expression to the same basis.

Knowing that $8=2^3$ and $1/2 = 2^-1$ we have

$2^(3(x-1)) le 2^(-(2x-1))$

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

$3(x-1) le -(2x-1)$ or

$5x-4 le 0$ so $x le 4/5$ is the feasible set

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