If A=1+r^a+r^ 2 a +r^ 3 a + ∞ and B=1+ ^ b + ^ 2 b + ^ 3 b + ∞ ,...

$A=\frac{1}{1-r^a} \Rightarrow 1-r^a=\frac{1}{A} \Rightarrow r^a=1-\frac{1}{A}=\frac{A-1}{A}B=\frac{1}{1-r^b} \Rightarrow 1-r^b=\frac{1}{B} \Rightarrow r^b=1-\frac{1}{B}=\frac{B-1}{B}\therefore a \log r=\log$\left(\frac{A-1}{A}\right)$and$b \log r=\log \left(\frac{B-1}{B}\right)\therefore \frac{\mathrm{a}}{\mathrm{b}}=\frac{\log \left(\frac{\mathrm{A}-1}{\mathrm{ A}}\right)}{\log\left(\frac{\mathrm{B}-1}{\mathrm{ B}}\right)}=\log _{\frac{\mathrm{B}-1}{\mathrm{ B}}} $\left(\frac{\mathrm{ A}-1}{\mathrm{ A}}\right)$