Question

Solutions set of the inequality: $3^x{\left(\mathrm{0.333...}\right)}^{(x+3)}\le {\left(\frac{1}{9}\right)}^x$ is

• A. $(\frac{3}{2},5)$
• B. $(-\infty ,\frac{3}{2}]$
• C. $(0,\infty )$
• D. none of these

Answer 1

The correct option is B $(-\infty ,\frac{3}{2}]$

Using $\mathrm{0.33...}=\frac{1}{3}$, we can write the inequality as $3^x{\left(\frac{1}{3}\right)}^{x+3}\le 3^{-2x}$
$=3^{-3}\le 3^{-2x}\iff -3\le -2x\iff x\le \frac{3}{2}$
or $x\in (-\infty ,\frac{3}{2}]$
Ans: B