Question

The value of $a^{{\log }_2}x$, where $a=0.2,b=\sqrt{5},x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....$ to $\infty$ is

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

Answer 1

Answer: (D)

$4$

$x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....,a=0.2,b\sqrt{5}x\frac{\frac{1}{4}}{1-\frac{1}{2}}x=\frac{1}{2}{a}^{{\log }_{b}x}=(0.2{)}^{{\log }_{\sqrt{5}}\frac{1}{2}}=(\frac{1}{2}{)}^{{\log }_{\sqrt{5}}0.2}=(\frac{1}{2}{)}^{-2}{a}^{{\log }_{b}x}=4$