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The value of (0.16)log2.5 (1/3 + 1/32 + 1/33 + ... to ∞ ) is equal to _______ . 

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\(\frac 13 + \frac1{3^2} + \frac 1{3^3} + ... \infty\)

\(= \frac{1/3}{1 - 1/3}\)

\(= \frac 12\)

\((0.16)^{\log_{2.5}(\frac 13 + \frac 1{3^2} + \frac 1{3^3} + ...\infty)}\)

\(= [(0.4)^2]^{\log_{\frac 52}\frac 12}\)

\(= (0.4)^{\log_{\frac 52}\frac 14}\)

\(= (\frac 14) ^{\log_{\frac 52}\frac 25}\)   \((\because a^{\log _bc} = c^{\log_ba})\)

\(= (4) ^{-\log_\frac 52 \frac 25}\)

\(= (4) ^{\log_\frac 25 \frac 25}\)

\(=4\)

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4

(0.16)log2.5 (1/3 + 1/32 + 1/33 + ... to ∞ )

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