The value of (0.16)^_2.5(13+(1)/(3^2)+(1)/(3^3)+) is equal to ________.

Question

The value of $(0.16)^{\log_{2.5}\left(\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots\infty\right)}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 4

Accepted Answer

$\dfrac13+\dfrac{1}{3^2}+\cdots = \dfrac12$, so the expression is $(0.16)^{\log_{2.5}(1/2)} = (2.5)^{-2\log_{2.5}(1/2)} = \left(\dfrac12\right)^{-2} = 4.$

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