The value of $(0.04^{log_{\sqrt{5}}(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...)})$ is __________.

The series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$ is an infinite GP with $a = \dfrac{1}{4}$ and $r = \dfrac{1}{2}$

Using the infinite GP sum formula:

$S_\infty = \dfrac{a}{1 - r} = \dfrac{1/4}{1 - 1/2} = \dfrac{1/4}{1/2} = \dfrac{1}{2}$

So our expression becomes:

$(0.04)^{\log_{\sqrt{5}}\left(\frac{1}{2}\right)}$

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Properties of logarithms we'll use:

$\text{P1: } \log_{n^b}(m^a) = \dfrac{a}{b} \log_n m \qquad \text{P2: } a^{\log_b c} = c^{\log_b a} \qquad \text{P3: } \log_a a = 1$

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Simplifying the logarithm using P1:

$\log_{\sqrt{5}}\left(\frac{1}{2}\right) = \log_{5^{1/2}}(2^{-1}) = \dfrac{-1}{1/2} \cdot \log_5 2 = -2\log_5 2$

The $-1$ (exponent of $2$) goes to the numerator, and $\frac{1}{2}$ (exponent of $5$) goes to the denominator, giving $\dfrac{-1}{1/2} = -2$

So we now have:

$(0.04)^{-2\log_5 2}$

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Converting $0.04$ to a cleaner form:

$0.04 = \dfrac{4}{100} = \dfrac{1}{25} = 25^{-1}$

Substituting:

$(25^{-1})^{-2\log_5 2}$

Multiplying the exponents (power raised to a power):

$= 25^{(-1) \times (-2\log_5 2)} = 25^{2\log_5 2}$

Since $2\log_5 2 = \log_5 2^2 = \log_5 4$:

$= 25^{\log_5 4}$

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Using P2 to swap — $a^{\log_b c} = c^{\log_b a}$:

$25^{\log_5 4} = 4^{\log_5 25}$

Since $\log_5 25 = \log_5 5^2 = 2$ (using P3):

$= 4^2 = \boxed{16}$