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?  


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*}
