Question

If $y={\left(0.64\right)}^{{\log }_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots \infty )}$, then the value of $y$ is equal to

• A. $0.25$
• B. $0.6$
• C. $0.8$
• D. $0.9$

Answer 1

Answer: (C)

$0.8$

Let $S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots \infty$

$\Rightarrow S=\frac{\frac{1}{3}}{1-\left(\frac{1}{3}\right)}=\frac{1}{2}$

$\Rightarrow {\log }_{0.25}S=\frac{{\log }_{0.5}\left(0.5\right)}{2}[\because {\log }_{b^n}{a}^m=\frac{m}{n}{\log }_{b}a]=\frac{1}{2}$

Therefore, $y={\left(0.64\right)}^{\frac{1}{2}}=0.8$
Ans: C