<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] Simplifying this equation by the step below:
$(2x^2+3)\log_{9}(2x+3)=(2x^2+3)\frac{\log_{3}(2x+3)}{\log_{3}9}=\frac{2x^2+3}{2}\log_{3}(2x+3)$
Now the equation is : $x^2\log_{3}x^2-\frac{2x^2+3}{2}\log_{3}(2x+3)=3\log_{3}\frac{x}{2x+3}$
Eliminate the log symbol, we get: $\frac{x^{2x^2}}{(2x+3)^{\frac{2x+3}{2}}}=(\frac{x}{2x+3})^3$
Reduce and simplify: $x^{2x^2-3}=(2x+3)^{\frac{2x^2-3}{2}}$
Now, we see $2x^2-3=0$ or $x=(2x+3)^{\frac{1}{2}}$, namely, $x^2-2x-3=0$
So the roots are $x_1=\sqrt{\frac{3}{2}}, x_2=-\sqrt{\frac{3}{2}},x_3=-1,x_4=3$
Note that log function requires positive values, so $x_2$ and $x_3$should be discarded by substituting them into the original equation.
Conclusion: the roots are $x_1=\sqrt{\frac{3}{2}}, x_2=3$
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