0-16-log-2-5-1-3-1-3-2-1-3-3-

Question Number 116491 by bemath last updated on 04/Oct/20

{(0.16)}^{\log_{2.5} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots)} = ?

Answered by bobhans last updated on 04/Oct/20

Only applying property of logarithm

\Rightarrow a^{\log_{a} f (x)} = f (x)

\Rightarrow {(\frac{2}{5})}^{2 . \log_{2.5} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots)} = {(\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots)}^{- 2}

= {(\frac{\frac{1}{3}}{1 - \frac{1}{3}})}^{- 2} = {(\frac{1}{2})}^{- 2} = {(2)}^{2} = 4

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