# How do you solve the exponential inequality 8^(x-1)<=32^(3x-2)?

Jun 19, 2018

$x \ge \frac{3}{4}$

#### Explanation:

${8}^{x - 1} \le {32}^{3 x - 2}$

change the $8 \text{ and } 32$ to powers of 2

${\left({2}^{3}\right)}^{x - 1} \le {\left({2}^{5}\right)}^{3 x - 2}$

ie

${2}^{3 x - 1} \le {2}^{15 x - 10}$

$\implies 3 x - 1 \le 15 x - 10$

solve for $x$

$- 1 + 10 \le 15 x - 3 x$

$9 \le 12 x$

$12 x \ge 9$

$\implies x \ge \frac{9}{12}$

$x \ge \frac{3}{4}$