Question

Let $x={\log }_{0.4}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\infty )$. The value of ${\left(0.16\right)}^x$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. $\frac{1}{16}$

Answer 1

Answer: (A)

$\frac{1}{4}$

$x={\log }_{0.4}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...\infty \right)$

$={\log }_{0.4}\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)={\log }_{0.4}\left(0.5\right)$

Now, $(0.16{)}^x=(0.16{)}^{{\log }_{0.4}\left(0.5\right)}$

$=(0.16{)}^{\frac{2}{2}{\log }_{0.4}\left(0.5\right)}$

$=(0.16{)}^{{\log }_{{\left(0.4\right)}^2}{\left(0.5\right)}^2}$

$=(0.16{)}^{{\log }_{0.16}{\left(0.5\right)}^2}$

$={\left(0.5\right)}^2=\frac{1}{4}$