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I have tried to solve this multiple ways, but I keep getting $2\log_{3}x$. According to the answer key, it is supposed to work out to 0, but I'm not seeing it. Can someone point me in the right direction?

$\log_{3} 9x^4 - log_{3}(3x)^2 $

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  • $\begingroup$ One term is $\log_3(9x^4)$, the other term is $\log_3(9x^2)$. They're clearly not equal ($x^4$ is not the same thing as $x^2$), so you can't get $0$ by subtracting them. This leaves three possibilities as I see it: 1) You haven't written down the problem correctly, or 2) You're looking at the answer key for the wrong problem, or 3) there is a mistake in the book. $\endgroup$ – Arthur Sep 2 '15 at 13:06
  • $\begingroup$ either the answer or the question is misprinted $\endgroup$ – David Quinn Sep 2 '15 at 13:06
  • $\begingroup$ I'm seeing the same result. $\log\left(\frac{9 x^4}{9x^2}\right) = \log x^2$. There's no getting around it. $\endgroup$ – user24142 Sep 2 '15 at 13:06
  • $\begingroup$ Thanks! I guess it's an issue with the answer key. I've checked for typos 5 times and there are none. I also checked the answer key over and over hoping it'd change, but it still says 0. Section 2, Number 1, e $\endgroup$ – Lisa Ever Sep 2 '15 at 13:10
  • $\begingroup$ It's not being evaluated or anything? Like at x=1? $\endgroup$ – snulty Sep 2 '15 at 13:20
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Use two properties: $$ \log_{a}{bc} = \log_{a}{b}+ \log_{a}{c}$$ and $$ \log_{a}{(b^c)} = c\cdot\log_{a}{b}.$$

And it seems it is typo and your answer is right.

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