How to solve $\log _{x^{2}-3}(x^{2}+6x)<\log _{x}(x+2)$?

Question

How to solve the following inequality $$\log _{x^{2}-3}\left(x^{2}+6x\right)<\log _{x}(x+2)\ ?$$

Answer 1

Note first that $x \in A \cup B$ where $A = (\sqrt3, 2), \; B = (2,\infty)$, for the question to make sense.

Now if $x \in A$, note that $0< x^2-3 < 1 < x < x+2 < x^2+6x$, so LHS $< 0$, while $RHS > 0$ so this is a solution.

For $x \in B$, we have $ 1 < x < x+2 $ and $1 < x^2-3 < x^2+6x$, so both LHS and RHS are positive. Here we have the equivalent inequality: $$\log_x (x+2) > \log_{x^2-3} (x^2+6x) \iff \log (x^2-3) \log(x+2) > \log x \log (x^2+6x)\\ \iff \log_x (x^2-3) > \log_{x+2} (x^2+6x)$$

Now note that $x^2 - 3 < x^2$ while $x^2+6x > (x+2)^2$ for $x \in B$.