$a = \log_{40}100, b = \log_{10}20$.How can I express $b$ depending only on $a$?

Question

Let $a = \log_{40}{100}, b = \log_{10}{20}$. How can I express $b$ depending only on $a$? I tried using the formula to change the base from $40$ to $10$, but couldn't get it just depending on $a$.

I used the base change formula $\log_a b = \dfrac{\log_ c b}{\log_c a}$ and got that $\log_{40} 100 = \dfrac{\log_{10} 100}{\log_{10}{40}} = \dfrac{2}{\log_{10}{20} + \log_{10}{2}}$. But then how could I express $\log_{10}2$ depending on $\log_{10}20$? I think that's what it suffices to show.

Answer 1

$$a=\log_{40}{100}=2\log_{40}10=\frac{2}{\log_{10}40}=\frac{2}{1+2\log_{10}2},$$ which gives $$\log_{10}2=\frac{2-a}{2a}.$$ Id est, $$\log_{10}20=1+\log_{10}2=\frac{a+2}{2a}.$$

Answer 2

Find $a=\log_{40}100$ only in terms of $b=\log_{10}20$.

\begin{align*} 2b&=2\log_{10}20=\log_{10}(20)^2=\log_{10}(40\cdot10)=\log_{10}40+1=2\log_{10^2}40+1=\frac2a+1 \end{align*}