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

by abcdmath   Last Updated October 13, 2019 09:20 AM - source

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

Tags : logarithms


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