Help on This Inequality

• February 24th 2010, 07:23 PM
Help on This Inequality
i cannot remember how to solve for this:
9^x^2<=27
• February 24th 2010, 08:11 PM
TheEmptySet
Quote:

Originally Posted by Zamadatix
i cannot remember how to solve for this:
9^x^2<=27

Since the expoential is positive and the log is an increasing function we can take the log of both sides of the equation.

If you use the base 3 log you get

$\log_3(9)^{x^2}=\log_3(27) \implies 2x^2 \le 3 \iff 2\left(x^2-\frac{3}{2}\right) \le 0$

$\left(x-\sqrt{\frac{3}{2}} \right)\left(x+\sqrt{\frac{3}{2}} \right)\le 0 \implies x \in \left[ -\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}\right]$