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Algebra Level 3

Let $A = 1+r^a + r^{2a} + r^{3a} + \cdots$ , and
let $B = 1+r^b + r^{2b} + r^{3b} + \cdots$ .

If $|r| < 1$ , which of the following is equal to $\dfrac ab$ ?

\dfrac{log (\dfrac{A-1}{A})}{log (\dfrac{B-1}{B})}

\dfrac{log{B}}{log{A}}

\dfrac{log{(1-A)}}{log{(1-B)}}

\dfrac{log{(1+A)}}{log{(1+B)}}

\dfrac{logA}{logB}

1 solution

Aditya Dhawan
May 1, 2016

$A=\quad \frac { 1 }{ 1-{ r }^{ a } } \quad and\quad B=\frac { 1 }{ 1-{ r }^{ b } } \\ \Rightarrow \frac { A-1 }{ A } =\quad { r }^{ a }\quad and\quad \frac { B-1 }{ B } =\quad { r }^{ b }\\ \Rightarrow \log { \frac { A-1 }{ A } = } log{ \quad r }^{ a }=\quad a\quad log\quad r\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad and\quad \\ \log { \frac { B-1 }{ B } = } log\quad { r }^{ b }=\quad b\quad log\quad r\\ \therefore \boxed {\frac { \log { \frac { A-1 }{ A } } }{ \log { \frac { B-1 }{ B } } } =\quad \frac { a }{ b } } \\$

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Simple standard approach.

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