$ 2\log ^2_{4}(|x+1|)+\log_4(|x^2-1|)+\log_{\frac{1}{4}}(|x-1|)=0$

Question

Find the sum of solutions to:

$$ 2\log^2_{4}(|x+1|)+\log_4(|x^2-1|)+\log_{\frac{1}{4}}(|x-1|)=0 $$

I'm not sure about what to do with the absolute values, how can I get rid of them?

Should I solve for all various cases depending on the sign of $x+1$ and $x-1$?

Answer 1

Note that $$ \log_{1/a}x=-\log_a x $$ so your equation becomes $$ 2(\log_4|x+1|)^2+\log_4|x+1|+\log_4|x-1|-\log_4|x-1|=0 $$ So $$ \log_4|x+1|\bigl(2\log_4|x+1|+1)=0 $$ Can you go on?

Answer 2

I am just adding the solution for you to check after your own computation, the last comment of David contains already the main idea:

So I got $x_1=-2,x_2=0,x_3=-3/2,x_4=-1/2$

bests