Solve $\ln(x^2 - 2x -2) \leq 0$

Question

I'm trying to solve the inequality $\ln(x^2 - 2x -2) \leq 0$ Just want to make sure that I'm doing it right.

$$\ln(x^2 - 2x -2) \leq 0$$

$x^2 - 2x -2 \leq e^0$ since $e^x$ is a strictly increasing function

$$x^2 - 2x - 3 \leq 0$$

$$(x+1) (x-3) \leq 0 $$

Therefore, the solutions are $-1\leq x\leq 3$

Please let me know if I did it correctly.

Thanks,

@SohamChowdhury It looks like he had that, then subtracted $1$ from both sides. — Mike, Oct 12 '14 at 05:09

score 6 · Answer 1 · answered Oct 12 '14 at 05:07

A little more needs to be done. Note that $x^2-2x-2=(x-1)^2-3$. If this is $\le 0$, then $\ln(x^2-2x-2)$ is not defined.

So you will have to examine your interval $[-1,3]$, and throw away the numbers in this interval for which the logarithm is not defined.

1 Answers1