$$3\log_{\frac{4}{9}}\sqrt[4]{\frac{27}{8}}$$
$$\log_{\frac{3}{2}}\frac{16}{81}$$
I understand using the expansion property to expand the division into a subtraction but how do I proceed from there?
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First observe that $$ 3\log_\frac{4}{9}\sqrt[4]{\frac{27}{8}} =\frac{3}{4}\log_\frac{4}{9}\frac{27}{8} $$ So all we need now is to compute the log term. By definition $\log_ab=y$ means that $a^y=b$ so we need to find $y$ such that $$ \left(\frac{4}{9}\right)^y=\frac{27}{8} $$ Expanding this, we have $$ \left(\frac{2^2}{3^2}\right)^y=\frac{3^3}{2^3} $$ or $$ \left(\frac{2}{3}\right)^{2y}=\left(\frac{3}{2}\right)^3 $$ and we're almost there, if we could make the left and right sides have the same base. Can you carry on from there?
Rick Decker
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The trick here is to not go to the subtraction step. Observe, for example, for the second expression that $\frac{16}{81} = (\frac{2}{3})^4 = (\frac{3}{2})^{-4}$, and hence the second logarithm is equal to $-4$. The first one is similar. Let me know if you have any difficulties.
Lord Soth
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Edit: it's correct.
– Kevin Li Dec 04 '13 at 19:53