If $A = 1 + r^{a} + r^{2a} + \cdots ~~~~~ ;~~~ B= 1 + r^{b} + r^{2b} + \cdots $
Then $a/b$ is ?
I took the sum of both series :
$$A = \dfrac{1}{1-r^a}~~;~~~ B=\dfrac{1}{1-r^b}$$
But now how do we break this to find $a/b$
Answer was given to be $\log_{\frac{B-1}{B}}(\frac{A-1}{A})$. How did they get this?
Hint
If $$A = \dfrac{1}{1-r^a}$$ $$r^a=\frac{A-1}A$$ Take logarithms of both sides to get $$a=\frac{\log \left(\frac{A-1}{A}\right)}{\log (r)}$$ Similarly $$b=\frac{\log \left(\frac{B-1}{B}\right)}{\log (r)}$$ Then $$\frac ab=\frac{\log \left(\frac{A-1}{A}\right)}{\log \left(\frac{B-1}{B}\right)}$$ Now change the base of the logarithms.