If $\log_{10}15=a$ and $\log_{20}50=b$, express $\log_940$ in terms of $a$ and $b$.

Question

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?

Answer 1

Guide:

$$\log_{20}50=\frac{\log_9 50}{\log_9 20}=\frac{2\log_95 + \log_92}{2\log_92+\log_95}=b$$

$$\log_{10}15 = \frac{\log_915}{\log_910}=\frac{\log_95+\log_93}{\log_92+\log_95}=a$$

Solve $\log_95$ and $\log_92$ in terms of $a$ and $b$.

$$\log_9 40 = \log_9 5 + 3 \log_9 2$$

Answer 2

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 $\log_{10}5 = 1-\log_{10}2$.