Solutions of $\log_{\frac{1}{4}}(x^2 - 1) < 2 \log_{\frac{1}{4}}(x-2)$?

Question

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 $x > 2$.

Answer 1

Actually it is pretty straightforward since we are dealing with the same basis, namely $\frac14$, within both logarithms. First of all utilizing the basic loagrithmic law $\log(a^b)=b\log(a)$ we get

$$\log_\frac14[x^2-1]<2\log_\frac14[x-2]\Rightarrow\log_\frac14[x^2-1]<\log_\frac14[(x-2)^2]$$

Note that we now got a always postitive square within the second logarithm whereas the original inequalitiy is restricited to $x>2$$($the first logarithm is restricted to $|x|>1$, but since $1<2$ it is sufficient to only rely on the second greater restriction$)$. Keeping this in mind we can now eventually solve the inequality

\begin{align} \log_\frac14[x^2-1]&<\log_\frac14[(x-2)^2]\\ x^2-1&\color{red}{>}(x-2)^2\\ x^2-1&>x^2-4x+4\\ 4x&>5\\ \therefore~\frac54&<x \end{align}

However, since $\frac54<2$ we still have to choose the larger value which overall leads to the the solution $x>2$.

---

Beside the fact that you have not recalled the initial condition you probably made a mistake while applying the exponentation. Since

$$\ln\left(\frac14\right)=-\ln(4)<0$$

you have to reverse the direction of the comparison sign.

Answer 2

Hint: we can write $$\frac{\ln(x^2-1)}{\ln(\frac{1}{4})}<\frac{2\ln(x-2)}{\ln(\frac{1}{4})}$$ and $$2\ln(x-2)=\ln(x-2)^2$$ Since $$\ln(\frac{1}{4})<0$$ you will get $$x^2-1>(x-2)^2$$

Answer 3

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 $$x>2.$$