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

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 $x-1$?

• you need to get rid of all the logs by using the change of base formula – David Quinn Jun 23 '15 at 15:46
• I am sorry, didn't write the first log right, will that be the case now? – gvidoje Jun 23 '15 at 15:48
• OK, but now use the change of base formula on the last term only and you can simplify and factorize the whole expression... – David Quinn Jun 23 '15 at 15:53
• the title and body show a totally different story – RE60K Jun 23 '15 at 16:16
• @egreg maybe that's not what he was wishing – RE60K Jun 23 '15 at 16:22

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?

• Yeah, I used substation the first time when solving, so I had t(2t+1)=0.. and only went on to solve for t=0, somehow I mixed up exponential and log equations so I thought that 2t+1>0 for all t values, but since it's a log it can also be 0, thanks for reminding me :D – gvidoje Jun 23 '15 at 16:36

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

• I get the first two solutions, bot not the third and forth :/ – gvidoje Jun 23 '15 at 16:25
• yeah...the first two are already the obvious ones, for the other ones you need to do some transformation and remember that $log(a*b)=log(a)+log(b)$...you can also post your calculations and we check them – user190080 Jun 23 '15 at 16:28
• and since there is a bunch of editing going on you better make clear which one is there version you really want to solve, I solved $2\log^2_{4}(|x+1|)+\log_4(|x^2-1|)+\log_{\frac{1}{4}}(|x-1|)==0$ – user190080 Jun 23 '15 at 16:30
• I edit it myself to be the correct version, it's a squared log, but since I'm a beginner at writing these, I made an error the first time. – gvidoje Jun 23 '15 at 16:32