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

Question

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?

Answer 1

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.