<s>[INST] If $\log_{10}15=a$ and $\log_{20}50=b$, express $\log_940$ in terms of $a$ and $b$. I've done this so far: $$\log_940=\frac{\log_{10}40}{\log_{10}9}=\frac{1+\log_{10}4}{a+\log_{10}6-1}.$$ How do I proceed?
 [/INST] Well, this will be little messy but here is an attempt:
$$b = \frac{\log_{10}50}{\log_{10}20} = \frac{1+\log_{10}5}{1+\log_{10}2} = \frac{2-\log_{10}2}{1+\log_{10}2} \implies \log_{10}2 = \frac{2-b}{1+b}$$
and
$$a = \log_{10}15 = \log_{10}3+\log_{10}5 = \log_{10}3 + 1 - \log_{10}2 \Rightarrow \log_{10}3 = a+\log_{10}2-1$$
and notice that
$$\log_{9}40 = \frac{\log_{10}40}{\log_{10}9} = \frac{1+2\log_{10}2}{2\log_{10}3}$$
By the above equations, you have $\log_{10}2$ and $\log_{10}3$ in terms of $a$ and $b$ so you can find the answer from that.
Also, note that in above equations, we used the fact that $\log_{10}5 = 1-\log_{10}2$.
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