Question

Find the value of $(0.04{)}^{{\log }_5(0.1+0.01+0.001+....)}$

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Answer 1

0.04 = 0.2 $\times$ 0.2 = $(\frac{1}{5}{)}^2$

${0.04}^{{\log }_5(0.1+0.01+0.001+....)}$

Given series 0.1 + 0.01 + 0.001 ........ is a GP.Sum of infinite term of GP = $\frac{a}{1-r}$

= $(\frac{4}{100}{)}^{{\log }_5\left(\left(\frac{0.1}{1-0.1}\right)\right)}$

= $5^{-2\times {\log }_5\left(\frac{1}{9}\right)}$

= $5^{{\log }_5\left(81\right)}$ {we know $a^{{\log }_{a}x}=x$}

= 81

Small proof for $a^{{\log }_{a}x}=x$