Expressing logarithms in terms of $a$ and $b$

Question

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?

Answer 1

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.

Answer 2

After playing around, I noticed a few things:

$log_{3}5=ab$

$log_{3}2=a-ab$

Maybe use these facts after converting $log_{48}65$ in base $3$?