<s>[INST] Logarithm Subtraction and Division with Same Bases I'm rusty on logarithms.  What is the approach to a problem like this?  Any hints would be appreciated. 
I'm thinking the subtraction on the numerator and denominator can become division since the bases are the same?  

 [/INST] Often better to factor the arguments of the logarithms to get to much simpler logarithms more quickly...  \begin{align*}
&\frac{\log_2 24 - \frac{1}{2} \log_2 72}{\log_3 18 - \frac{1}{3} \log_3 72}  \\
    &\quad{}= \frac{\log_2 (2^3 \cdot 3) - \frac{1}{2} \log_2 (2^3 \cdot 3^2)}{\log_3(2\cdot 3^2) - \frac{1}{3} \log_3 (2^3 \cdot 3^2)}  \\
    &\quad{}= \frac{\log_2 2^3 + \log_2 3 - \frac{1}{2} \left( \log_2 2^3 + \log_2 3^2 \right)}{\log_3 2 + \log_3 3^2 - \frac{1}{3} \left( \log_3 2^3 + \log_3 3^2 \right)}  \\
    &\quad{}= \frac{3\log_2 2 + \log_2 3 - \frac{1}{2} \left( 3\log_2 2 + 2\log_2 3 \right)}{\log_3 2 + 2\log_3 3 - \frac{1}{3} \left( 3\log_3 2 + 2\log_3 3 \right)}  \\
    &\quad{}= \frac{3 + \log_2 3 - \frac{1}{2} \left( 3 + 2\log_2 3 \right)}{\log_3 2 + 2 - \frac{1}{3} \left( 3\log_3 2 + 2 \right)}  \\
    &\quad{}= \frac{3 + \log_2 3 - \frac{3}{2} - \log_2 3 }{\log_3 2 + 2 - \log_3 2 - \frac{2}{3}}  \\
    &\quad{}= \frac{3 - \frac{3}{2}}{2 - \frac{2}{3}} \cdot \frac{6}{6}  \\
    &\quad{}= \frac{18-9}{12-4}  \\
    &\quad{}= \frac{9}{8}  \text{.}
\end{align*}
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