As in the title, what are the solutions to $$\log_{\frac{1}{4}}(x^2 – 1) < 2 \log_{\frac{1}{4}}(x-2)$$ ?
I have tried to use the logarithm's power and subtraction rules, but it seems like the answer $x < \frac{5}{4}$ I came up with is wrong. Any tips on this? Really can't figure out what I'm doing wrong; how would you guys go about solving this? Thank you.
Update
This is what I had done thus far (seems like it was cropped out…):
$$\log_{\frac{1}{4}}(x^2 – 1) < \log_{\frac{1}{4}}(x – 2)^2$$
$$x^2 – 1 > (x – 2)^2$$
$$1 + 4 < 4x, x > \frac{5}{4}$$
and mistaken this condition as the solution without considering the inequality's overall domain and thus $x > 2$.
Best Answer
Since $(0,\infty)\ni x\mapsto \log_{\frac{1}{4}}x\in \Bbb R$ is strictly decreasing, $$ \log_{\frac{1}{4}}(x^2 - 1) < 2 \log_{\frac{1}{4}}(x-2)= \log_{\frac{1}{4}}(x-2)^2 $$ if and only if $$ x^2 - 1>(x-2)^2=x^2-4x+4, $$ and $$x^2-1>0,\quad x-2>0.$$ If we solve it, we get $4x>5$, $|x|>1$ and $x>2$. Therefore the solution is $$x>2.$$