<s>[INST] Expressing logarithms in terms of $a$ and $b$ Given that $\log_{10}{5}=a$ and $\log_{3}{10}=b$, express $\log_{48}{65}$ in terms of $a$ and $b$.
I tried various transformations of $\log_{48}{65}$ and got:
$$\log_{48}{65} =\frac{\log_{10}{65}}{\log_{10}{48}}=\frac{a+\log_{10}{13}}{\frac{1}{b}+4\log_{10}2}=\frac{a+\log_{10}{13}}{\frac{1}{b}+4-4a}$$
I'm stuck here since I'm really not sure what to do with $\log_{10}{13}$. Have I approached this problem in a wrong way, and if so, what is a better way to approach it?
 [/INST] I'd just put all in terms of $\log_{10} = \log$ (just for familiarity with days of yore):
$\begin{align*}
  \log 5
    &= a \\
  \log 2
    &= \log 10 - \log 5 \\
    &= 1 - a \\
  \log_3 10
    &= \frac{\log 10}{\log 3} \\
    &= \frac{1}{\log 3} \\
    &= b \\
  \log_{48} 65
    &= \frac{\log 65}{\log 48} \\
    &= \frac{\log 5 + \log 13}{\log 3 + \log 16} \\
    &= \frac{(a + \log 13)}{1/b + \log 16} \\
    &= \frac{b (a + \log 13)}{1 + 4 b \log 2} \\ 
    &= \frac{b (a + \log 13)}{1 + 4 b (1 - a)}
 \end{align*}$
Can't get rid of the 13, as a prime it can't be written as a product of powers of 10 and 3.
</s>