# How do you solve 5^(x+3) <= 2^(x+4)?

Aug 9, 2015

Replace $5$ with ${e}^{\ln \left(5\right)}$ and $2$ with ${e}^{\ln \left(2\right)}$; then solve using the new exponents (ignoring the identical base values):
$\textcolor{w h i t e}{\text{XXXX}}$$x \le - 2.24353$

#### Explanation:

Let $k = \ln \left(5\right) = 1.609438$ and
let $m = \ln \left(2\right) = 0.693147$

$5 = {e}^{k}$$\textcolor{w h i t e}{\text{XXXX}}$and$\textcolor{w h i t e}{\text{XXXX}}$$2 = {e}^{m}$

${5}^{x + 3} \le {2}^{x + 4}$
$\textcolor{w h i t e}{\text{XXXX}}$is equivalent to
${\left({e}^{k}\right)}^{x + 3} \le {\left({e}^{m}\right)}^{x + 4}$

$k x + 3 k \le m x + 4 m$

$\left(k - m\right) x \le \left(4 m - 3 k\right)$
$\textcolor{w h i t e}{\text{XXXX}}$since $k > m$ we can divide both sides by $k - m$
$\textcolor{w h i t e}{\text{XXXX}}$without changing the orientation of the inequality

$x \le \frac{4 m - 3 k}{k - m}$

Substituting the value of $\ln \left(5\right)$ for $k$
and the value of $\ln \left(2\right)$ for $m$
then pushing it all through my calculator...

I get $x \le - 2.24353$