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?  

 A: 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*}
A: 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*}
A: Rule 1: $\ \log_a(x^q) = q\log_a(x)$
Rule 2 (Addition): $\ \log_a(x) + \log_a(y) = \log_a(xy), \ $ 
Rule 3 (Subtraction): $\ \log_a(x) - \log_a(y) = \log_a(\frac{x}{y}), \ $
Rule 3 is actually Rule 2 but replacing y with $\  \frac{1}{y} =y^{-1}$, and then using Rule 1.
Rule 4 (Change of base): $\ \frac{\log_a(x^p)}{\log_a(x^q)} = \log_q(p)$
A: Yes, the subtractions can become division, and the fractions can become roots, plus $\log_a b =\frac 1 {\log_b a}$
