# Solve for $x$: $\frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2} = \frac{5}{12}.$ [duplicate]

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?

• 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 $x > -2$ – Sil Oct 13 at 9:14
• Why do you think there is a closed form answer? – GEdgar Oct 13 at 9:21
• 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...) – Sil Oct 13 at 9:23
• @ GEdgar. There my not be. But I asked if there any. – abcdmath Oct 13 at 9:24
• Here (with natural log) the answer is approximately $x=9.43973090000793458249674162201$. As noted, it does have a nice answer for log base 2. – GEdgar Oct 13 at 9:30

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 $$0 and $$1 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

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

and for $$1

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

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.