# How do you solve the exponential inequality 5^(2x+2)>=25^(2x-3)?

Apr 14, 2017

$x \le 4$

#### Explanation:

Note that $25 = {5}^{2}$
So ${25}^{2 x - 3} = {\left({5}^{2}\right)}^{2 x - 3} = {5}^{4 x - 6}$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {5}^{2 x + 2} \ge {25}^{2 x - 3}$
is equivalent to
$\textcolor{w h i t e}{\text{XXX}} {5}^{2 x + 2} \ge {5}^{4 x - 6}$

which will be true if
$\textcolor{w h i t e}{\text{XXX}} 2 x + 2 \ge 4 x - 6$

Subtracting $2 x$ from both sides then adding $6$ (to both sides)
$\textcolor{w h i t e}{\text{XXX}} 8 \ge 2 x$

or
$\textcolor{w h i t e}{\text{XXX}} x \le 4$