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Solve for $x$: $$\frac{1}{\log\big((x+2)^2\big)}+\frac{1}{\log\big((x-2)^2\big)} = \frac{5}{12}.$$ My Attempt: \begin{align*} & \frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2} = \frac{5}{12} \\ \implies &\> \frac{1}{2\log(x+2)}+\frac{1}{2\log(x-2)} = \frac{5}{12} \\ \implies & \> \frac{1}{\log(x+2)}+\frac{1}{\log(x-2)} = \frac{5}{6} \\ \implies & \> 6\log(x^2-4) = 5 \log(x-2) \log(x+2).\end{align*} Please help me how can I proceed from here?

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    $\begingroup$ You need to be careful with $\log((x+2)^2)=2\log(x+2)$, it holds only when both sides are defined, but first expression is defined for $x \neq -2$, while second for $x > -2$ $\endgroup$ – Sil Oct 13 at 9:14
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    $\begingroup$ Why do you think there is a closed form answer? $\endgroup$ – GEdgar Oct 13 at 9:21
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    $\begingroup$ Almost a duplicate: Lograthmic equation $ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}$ solutions (it has base $2$, maybe that is what was intended? It has a nice solution then...) $\endgroup$ – Sil Oct 13 at 9:23
  • $\begingroup$ @ GEdgar. There my not be. But I asked if there any. $\endgroup$ – abcdmath Oct 13 at 9:24
  • $\begingroup$ Here (with natural log) the answer is approximately $x=9.43973090000793458249674162201$. As noted, it does have a nice answer for log base 2. $\endgroup$ – GEdgar Oct 13 at 9:30
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We have that

$$f(x)=\frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2}$$

is even and therefore we can assume $x> 0$ with $x\neq 1$ and $x\neq 2$ ($x=0$ is not a solution).

For $0<x<1$ and $1<x<2$ we have that

$$f(x)=\frac{1}{2\log(x+2)}+\frac{1}{2\log(2-x)}\\\implies f'(x)=\frac{1}{2(2-x)\log^2(2-x)}-\frac{1}{2(x+2)\log^2(x+2)}$$

from whic we can conlcude that for $0<x<1$

$$f(x)> \frac2{\log 4}$$

and for $1<x<2$

$$f(x)< \frac1{\log 16}$$

For $x>2$ we have

$$f(x)=\frac{1}{2\log(x+2)}+\frac{1}{2\log(x-2)}\\\implies f'(x)=-\frac{1}{2(x+2)\log^2(x+2)}-\frac{1}{2(x-2)\log^2(x-2)}<0$$

therefore on that interval $f(x)$ is strictly decreasing and since

  • $\lim_{x\to 2^+} f(x)=\infty$
  • $\lim_{x\to \infty} f(x)=0$

by IVT exactly a real solution exists which can be determined numerically and leads to $x\approx 11.3467$. For symmetry also $x\approx -11.3467$ is a solution.

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