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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 $x-1$?

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

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  • $\begingroup$ 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 $\endgroup$ – gvidoje Jun 23 '15 at 16:36
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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

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  • $\begingroup$ I get the first two solutions, bot not the third and forth :/ $\endgroup$ – gvidoje Jun 23 '15 at 16:25
  • $\begingroup$ 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 $\endgroup$ – user190080 Jun 23 '15 at 16:28
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    $\begingroup$ 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 $2\log^2_{4}(|x+1|)+\log_4(|x^2-1|)+\log_{\frac{1}{4}}(|x-1|)==0$ $\endgroup$ – user190080 Jun 23 '15 at 16:30
  • $\begingroup$ 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. $\endgroup$ – gvidoje Jun 23 '15 at 16:32

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