# How to solve log_(x^2)−3(x^2+6x)<logx(x+2)?

How to solve ${\mathrm{log}}_{{x}^{2}-3}\left({x}^{2}+6x\right)<{\mathrm{log}}_{x}\left(x+2\right)$?
How to solve the following inequality
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Note first that $x\in A\cup B$ where $A=\left(\sqrt{3},2\right),\phantom{\rule{thickmathspace}{0ex}}B=\left(2,\mathrm{\infty }\right)$, for the question to make sense.
Now if $x\in A$, note that $0<{x}^{2}-3<1, so LHS $<0$, while $RHS>0$ so this is a solution.
For $x\in B$, we have $1 and $1<{x}^{2}-3<{x}^{2}+6x$, so both LHS and RHS are positive. Here we have the equivalent inequality:
${\mathrm{log}}_{x}\left(x+2\right)>{\mathrm{log}}_{{x}^{2}-3}\left({x}^{2}+6x\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}\left({x}^{2}-3\right)\mathrm{log}\left(x+2\right)>\mathrm{log}x\mathrm{log}\left({x}^{2}+6x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}\left({x}^{2}-3\right)>{\mathrm{log}}_{x+2}\left({x}^{2}+6x\right)$
Now note that ${x}^{2}-3<{x}^{2}$ while ${x}^{2}+6x>\left(x+2{\right)}^{2}$ for $x\in B$