Question

If A = 1 + r^a + r^2a + r^3a + ..... ∞, B = 1 + r^b + r^2b + r^3b + ..... ∞, then show that \(\frac{a}{b}\) = \(\text{log}_{\big(\frac{B-1}{B}\big)}\)\({\big(\frac{A-1}{A}\big)}\).

Answer 1

A = 1 + r^a + r^2a + r^3a + ..... ∞

⇒ A = \(\frac{a}{1-r^a}\)      \(\big(\because\,S_\infty = \frac{a}{1-r}.\text{Here}\,a=1, \text{common ratio}=r^a\big)\)

⇒ 1 - r^a = \(\frac{1}{A}\) ⇒ r^a = 1 - \(\frac{1}{A}\)= \(\frac{A-1}{A}\) ⇒ a log r = log \(\big(\frac{A-1}{A}\big)\)

⇒ a = \(\frac{\text{log}\big(\frac{A-1}{A}\big)}{\text{log}\,r}\)                        .....(i)

Now,  B = 1 + r^b + r^2b + r^3b + .... ∞

 ⇒ B = \(\frac{a}{1-r^b}\)  ⇒ 1 - r^b = \(\frac{1}{B}\) ⇒ r^b = 1 - \(\frac{1}{B}\)= \(\frac{B-1}{B}\) ⇒ log r^b = log \(\big(\frac{B-1}{B}\big)\)

⇒ b log r = log \(\big(\frac{B-1}{B}\big)\) ⇒ b = \(\frac{\text{log}\big(\frac{B-1}{B}\big)}{\text{log}\,r}\)                .....(ii)

∴ \(\frac{a}{b}\) = \(\frac{\text{log}\big(\frac{A-1}{A}\big)}{\text{log}\,r}\) x \(\frac{\text{log}\,r}{\text{log}\big(\frac{B-1}{B}\big)}\)                  (From (i) and (ii))

⇒ \(\frac{a}{b}\) = \(\frac{\text{log}\big(\frac{A-1}{A}\big)}{\text{log}\big(\frac{B-1}{B}\big)}\) = \(\text{log}_{\big(\frac{B-1}{B}\big)}\)\({\big(\frac{A-1}{A}\big)}\).              \(\bigg[\because\frac{\text{log}\,a}{\text{log}\,b}\) = log_b a\(\bigg].\)