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To solve for x in $$3\log_3|-x| = \log_3 x^2 $$

I made cases such that when $x > 0$ and $x<0$

For first part i.e $x >0$ i get $3\log_3 x=2\log x^2$. so $x=1$.

For other case also comes out to be $x=1$. But textbook states answer to be $x=-1 $ also. Now clearly this satisfies eqiation also. But why this solution is not coming in my working method?

Thanks

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It should. When $x<0$ we have $|-x| = -x$ so

$$3 \log_3 |-x| = \log_3 x^2 \implies -x^3 = x^2 \implies x^2 + x^3 = 0 \implies x^2(x+1) = 0$$

so either $x=0$ which isn't in the domain, or $x+1 = 0 \implies x=-1$.

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  • $\begingroup$ Isn't that $\log(-x)$ is not defined? so we reject cases where x<0 $\endgroup$ – J. Deff Oct 22 '16 at 1:41
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    $\begingroup$ How? When $x<0$ then $-x > 0$ so $\log (-x)$ is perfectly well defined. The argument of your logarithm has to be positive, when $x<0$, the argument $-x$ is indeed positive. $\endgroup$ – Zain Patel Oct 22 '16 at 1:42
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    $\begingroup$ Ah thanks i got it $\endgroup$ – J. Deff Oct 22 '16 at 1:43
  • $\begingroup$ One question: was my question not perfectly typed since you have edited it $\endgroup$ – J. Deff Oct 22 '16 at 1:43
  • $\begingroup$ Sorry yeah, I just made the title more descriptive, latexified the "x=1" and used $\log$ instead of $log$ to make it look nicer - nothing of import - your question was well posed. :) $\endgroup$ – Zain Patel Oct 22 '16 at 1:44
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$$3\log_3|-x| = \log_3 x^2 $$

is equivalent to

$$3\log_3|x| = \log_3 |x|^2 $$

$$3\log_3|x| = 2\log_3 |x| $$

$$\log_3|x|=0$$

$$|x|=1$$

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Or the absolute value can be rescued:

$2 \cdot \frac{3}{2} \log_3 |x|=\log_3x^2 \Rightarrow \frac{3}{2} \log_3x^2=\log_3x^2 \Rightarrow \log_3x^2=0 \Rightarrow x^2=1 \Rightarrow x=\pm1.$

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