An algebra problem by Aly Ahmed

Algebra Level 2

A = 1 2 3 + 1 2 6 + 1 2 9 + ⋯ + 1 2 3 n + ⋯ , B = log ⁡ 128 2 \large A = \dfrac 1{2^3} + \dfrac 1{2^6} + \dfrac 1{2^9} + \cdots + \dfrac1{2^{3n}} + \cdots , \quad \quad B = \log_{128} 2

Find B log ⁡ A 4 B^{\log_A 4} .


The answer is 4.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Chew-Seong Cheong
Oct 10, 2016

A = 1 2 3 + 1 2 6 + 1 2 9 + . . . = 1 8 ( 1 1 − 1 8 ) = 1 7 \begin{aligned} A & = \frac 1{2^3} + \frac 1{2^6} + \frac 1{2^9} + ... \\ & = \frac 18 \left(\frac 1{1-\frac 18} \right) \\ & = \frac 17 \end{aligned}

B log ⁡ A 4 = ( log ⁡ 128 2 ) log ⁡ 1 7 4 = ( log ⁡ 128 12 8 1 7 ) log ⁡ 1 7 4 = ( 1 7 ) log ⁡ 1 7 4 = 4 \begin{aligned} B^{\log_A 4} & = (\log_{128} 2)^{\log_\frac 17 4} \\ & = (\log_{128} 128^\frac 17)^{\log_\frac 17 4} \\ & = \left(\frac 17\right)^{\log_\frac 17 4} \\ & = \boxed{4} \end{aligned}

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