# How is $x=-1$ a solution to $3\log_3|-x| = \log_3 x^2$?

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

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 $x+1 = 0 \implies x=-1$.

• Isn't that $\log(-x)$ is not defined? so we reject cases where x<0 – J. Deff Oct 22 '16 at 1:41
• 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. – Zain Patel Oct 22 '16 at 1:42
• Ah thanks i got it – J. Deff Oct 22 '16 at 1:43
• One question: was my question not perfectly typed since you have edited it – J. Deff Oct 22 '16 at 1:43
• 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. :) – Zain Patel Oct 22 '16 at 1:44

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

is equivalent to

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

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

$$\log_3|x|=0$$

$$|x|=1$$

Or the absolute value can be rescued:

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