Question

(0.05)^log√20(0.1+0.01+0.001+...)=​

Answer 1

Answer:

81

Step-by-step explanation:

Notice that 0.1 + 0.01 + 0.001 +...forms a inifite GP with sum a/(1 - r)= 0.1/(1 - 0.1) = 1/9

Thus,

$\implies \sf{(0.05) {}^{ log_{ \sqrt{20} } ( \frac{1}{9}) \: } } \\ \\ \implies\sf{( \small{\frac{1}{20}} ) {}^{ log_{20 {}^{ \frac{1}{2} } } ( \frac{1}{9}) \: } } \\ \\\implies \sf{( \small{20 {}^{ - 1} } ) {}^{ log_{20 {}^{ \frac{1}{2} } } ( 9 {}^{ - 1} ) \: } } \\ \\ \implies\sf{( \small{20 {}} ) {}^{ -1 \times ( - 1)log_{20 {}^{ \frac{1}{2} } } ( 9 {}^{} ) \: } } \\ \\\implies \sf{( \small{20 {}^{} } ) {}^{ log_{20 {}^{ \frac{1}{2} } } ( 9 {}^{ }) \: } } \\ \\ \implies\sf{( 9){}^{ log_{20 {}^{ \frac{1}{2} } } ( 20{}^{ } ) \: } } \\ \\\implies 9 {}^{2} \\ \\ \implies81$

Properties used :

Sum of infinite GP = a/(1 - r), starting

a^(logb) = b^(loga) , last 4th line