# How to Solve Logarithmic Inequalities

When you solve logarithmic inequalities, you use the same rules for equations as you are used to.

Rule

### Logarithmicinequalities

aking the exponent on both sides of an inequality does not change the inequality. Thus, nothing happens to the inequality sign! $\begin{array}{llllllll}\hfill \mathrm{log}x&

Example 1

Solve the inequality $\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)\le 8$ $\begin{array}{llll}\hfill \mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)& \le 8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)}& \le {e}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-4& \le {e}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le {e}^{8}+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \text{}2984.96\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Solve the inequality $\mathrm{log}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)\ge 4$

$\begin{array}{llllll}\hfill \mathrm{log}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)& \ge 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)}& \ge {e}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x+3& \ge {e}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x& \ge {e}^{4}-3\phantom{\rule{2em}{0ex}}& \hfill & |÷\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \frac{{e}^{4}-3}{-2}=\frac{3-{e}^{4}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ You need to flip the inequality sign when you divide by a negative number. You must have an equality sign in the last row as the answer is just changed to look nicer.

Example 3

Solve the inequality $\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\le e$ $\begin{array}{llll}\hfill \mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)& \le e\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}& \le {e}^{e}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-3& \le {e}^{e}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le {e}^{e}+3\approx \text{}18.15\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$