<s>[INST] Any idea how to solve this equation? Any idea how to solve this equation?
 $$x^2\log_{3}x^2-(2x^2+3)\log_{9}(2x+3)=3\log_{3}\frac{x}{2x+3}$$
 [/INST] Note that $\log_9(x) = \log_3(x)/\log_3(9) = \frac{1}{2}\log_3(x)$.
We need $2x+3\gt 0$, and hence $x\gt 0$, so that all logarithms can be evaluated. 
So we can simplify the equation using the basic properties of the logarithms 
$$\begin{align*}
x^2\log_{3}x^2-(2x^2+3)\log_{9}(2x+3) &= 3\log_{3}\left(\frac{x}{2x+3}\right)\\
2x^2\log_3(x) - \frac{2x^2+3}{2}\log_3(2x+3) &= 3\log_3(x) - 3\log_3(2x+3)\\
(2x^2-3)\log_3(x) &= \left(\frac{2x^2+3}{2} - 3\right)\log_3(2x+3)\\
(2x^2-3)\log_3(x) &= \frac{2x^2-3}{2}\log_3(2x+3).
\end{align*}$$
From this, it is clear that any solutions to $2x^2-3=0$ will satisfy the equation. The solutions are $x=\sqrt{3/2}$ and $x=-\sqrt{3/2}$; but the latter cannot be used in the original equation. So one possibility is
$$x = \sqrt{\frac{3}{2}}.$$
If $2x^2-3\neq 0$, then cancelling we get:
$$\begin{align*}
(2x^2-3)\log_3(x) &= \frac{2x^2-3}{2}\log_3(2x+3)\\
\log_3(x) &= \frac{1}{2}\log_3(2x+3)\\
2\log_3(x) - \log_3(2x+3) &= 0\\
\log_3\left(\frac{x^2}{2x+3}\right) &= 0\\
\frac{x^2}{2x+3} &= 1\\
x^2 -2x - 3 &= 0\\
(x-3)(x+1) &=0.
\end{align*}$$
Again, $x=-1$ cannot be used. So the only other solution is $x=3$.
So the solutions are $x=3$ and $x=\sqrt{\frac{3}{2}}$. 
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