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

$${(0.16)}^{\log {}_{2.5}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...)}=?$$

Answered by bobhans last updated on 04/Oct/20

$$\mathrm{Only}\mathrm{applying}\mathrm{property}\mathrm{of}\mathrm{logarithm}$$$$\Rightarrow {\mathrm{a}}^{\log {}_{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)}=\mathrm{f}\left(\mathrm{x}\right)$$$$\Rightarrow {\left(\frac{2}{5}\right)}^{2.\log {}_{2.5}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...)}={(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...)}^{-2}$$$$={\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)}^{-2}={\left(\frac{1}{2}\right)}^{-2}={\left(2\right)}^2=4$$

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