Help on This Inequality

**Zamadatix** · February 24th 2010, 07:23 PM

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

**TheEmptySet** · February 24th 2010, 08:11 PM

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

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

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

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