Question

Value of $y=(0.36{)}^{{\log }_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...\infty )}$

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

Answer 1

The correct option is C $0.6$

$Expression={0.36}^{lo{g}_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...)}$

$Expression={0.36}^{lo{g}_{0.25}\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)}$

$Expression={0.36}^{lo{g}_{0.25}\left(\frac{\frac{1}{3}}{\frac{2}{3}}\right)}$

$Expression={0.36}^{lo{g}_{0.25}\left(\frac{1}{2}\right)}$

$Expression={0.36}^{\frac{1}{2}}$

$Expression=0.6$