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

$$\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+....=\dfrac{\frac{1}{4}}{1-\frac{1}{2}}=\dfrac{\frac{1}{4}}{\frac{1}{2}}=\dfrac{1}{2}$$  (Applying the formula for infinite GP)

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=0.04^{\log_{\sqrt{5}}\left(\frac{1}{2}\right)}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=5^{\left(-2\right)\log_{\sqrt{5}}\left(\frac{1}{2}\right)}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=5^{-\dfrac{2}{\frac{1}{2}}\log_5\left(2\right)^{-1}}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=5^{4\log_52}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=2^{4\log_55}$$

=> $$0.04^{\log_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....)}=16$$