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Solve in $\mathbb{R}$: $$\log_3 x^2 - 2\log_{-x} 9 = 2$$

First, I put the condition that $x < 0$. Then, I've rewritten $\log_3 x^2$ as $2\frac{\log_{-x} x}{\log_{-x} 3}$ and got the following equation: $$\log_{-x} \frac{x}{3} = \log_{-x}^2 9$$

Now I need to solve this new equation, but I don't know how.

Thank you in advance!

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HINT:

For $x<0$

$\log_3x^2=2\log_3(-x)=y$(say)

and $\log_{-x}3=\dfrac{\log3}{\log(-x)}=\dfrac1{\dfrac{\log(-x)}{\log3}}=\dfrac1{\log_3(-x)}=\dfrac1y$

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  • $\begingroup$ solving quadratic you will get $x=-9$ & $x=-\frac13$ $\endgroup$ – Bhaskara-III Oct 20 '16 at 15:54

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