<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] The identities $\log_a b + \log_a c \equiv \log_a(bc)$ and
$b\log_a c \equiv \log_a c^b$ will be needed here. You can then do
\begin{align*}
 \frac{\log_2 24 - \frac 12 \log_2 72}
      {\log_3 18 - \frac 13 \log_3 72}
 &=
 \frac{\log_2 24 - \log_2 \sqrt{72}}
      {\log_3 18 - \log_3 \sqrt[3]{72}} \\
 &=
 \frac{\log_2 \frac{24}{\sqrt{72}}}
      {\log_3 \frac{18}{\sqrt[3]{72}}} \\
 &=
 \frac{\log_2 \frac{24}{3\sqrt 8}}
      {\log_3 \frac{18}{2\sqrt[3]{9}}} \\
 &=
 \frac{\log_2 \frac{8}{\sqrt 8}}
      {\log_3 \frac{9}{\sqrt[3]{9}}} \\
 &=
 \frac{\log_2 8 - \log_2 \sqrt 8}
      {\log_3 9 - \log_3 \sqrt[3]{9}} \\
 &=
 \frac{\log_2 2^3 - \log_2 2^{3/2}}
      {\log_3 3^2 - \log_3 3^{2/3}} \\
 &=
 \frac{3 - \frac 32}
      {2 - \frac 23} \quad \text{by definition} \\
 &= \frac 98
\end{align*}
</s>