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I'm trying to solve this logarithmic expression:

$\left(\dfrac{\left(\log_6(27)+2\log_6(2)\right)}{\log_6\sqrt[3]{0.25}+\log_6(\frac{1}{3})}\right)^{\!2}$

But after a couple of steps, I got stuck. Here's what I did: First, I simplified the numerator to $\log_6(108)$:

$\log_6(27)+2\log_6(2)$

$\log_6(27)+\log_6(4)$

$\log_6(27\cdot 4)=\log_6(108)$

In the denominator, I came as far as $-\frac{2}{3}\log_6(2)+\log_6(3)$:

  • $\log_6\sqrt[3]{0.25}=\frac{1}{3}\log_6(\frac{1}{4})=\frac{1}{3}\log_61-\frac{1}{3}\log_6(4)$ and since $\log_b(1)=0$, $\frac{1}{3}\log_6(1)$ cancels out. The remaining $-\frac{1}{3}\log_6(4)=-\frac{1}{3}\log_6(2)^2=-\frac{2}{3}\log_6(2)$
  • $\log_6(\frac{1}{3})=\log_6(1)-\log_6(3).$ Again, since $\log_b(1)=0$, $\log_6(1)$ cancels out. Which leaves $\log_6(3)$

So the denominator is now $-\frac{2}{3}\log_6(2)+\log_6(3)$

Have I been solving this problem correctly so far? If so, how should I continue the rest of it? I'm not sure whether I can use the logarithm product rule in $-\frac{2}{3}\log_6(2)+\log_6(3)$ like this: $-\frac{2}{3}\log_6(2)+\log_6(3)=-\frac{2}{3}\log_6(2)\cdot(3)$ since $\log_6(3)$ doesn't have the coefficient $-\frac{2}{3}$

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    $\begingroup$ The second term should be $-\log_6(3)$ $\endgroup$ – Shirish Kulhari Jul 26 '18 at 10:05
  • $\begingroup$ What do you call solving an expression? $\endgroup$ – Bernard Jul 26 '18 at 10:53
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Denominator = $$-\frac{1}{3}\log_6(4)-\frac{1}{3}\log_6(27) = -\frac{1}{3}\big(\log_6(4)+\log_6(27)\big) = -\frac{1}{3}\log_6(108)$$ Log term cancels out with numerator.

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You can simplify the computation: $$ \log_627+2\log_62=\log_627+\log_64=\log_6(27\cdot4)=\log_6108 $$ as you computed. The denominator is $$ \log_6\sqrt[3]{0.25}+\log_6\frac{1}{3}= \log_6\left(\sqrt[3]{0.25}\cdot\frac{1}{3}\right)= \log_6\sqrt[3]{\frac{25}{100}\cdot\frac{1}{27}}= \log_6\sqrt[3]{\frac{1}{108}}=-\frac{1}{3}\log_6108 $$

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