Question

The solution set of the equation ${\left[4\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+...\right)\right]}^{{\log }_{2}x}$ $={\left[54\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...\right)\right]}^{{\log }_{x}2}$ is:

• A. $\left\{4,\frac{1}{4}\right\}$
• B. $\left\{2,\frac{1}{2}\right\}$
• C. $\{1,2\}$
• D. $\left\{8,\frac{1}{8}\right\}$

Answer 1

Answer: (A)

$\therefore {\left[4\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\dots \right)\right]}^{{\log }_{2}x}$
$={\left[54\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\dots \right)\right]}^{{\log }_{x}2}$
$\Rightarrow {\left[4\left(\frac{1}{1+\frac{1}{3}}\right)\right]}^{{\log }_{2}x}={\left[44\left(\frac{1}{1-\frac{1}{3}}\right)\right]}^{{\log }_{x}2}$
$\Rightarrow {\left[4\left(\frac{3}{4}\right)\right]}^{{\log }_{2}x}={\left[54\times \frac{3}{2}\right]}^{{\log }_{x}2}$
$\Rightarrow 3^{{\log }_{2}x}=(81{)}^{{\log }_{x}2}$
$\Rightarrow 3^{{\log }_{2}x}=3^{4{\log }_{x}2}$
$\Rightarrow {\log }_{2}x=4{\log }_{x}2$
$\Rightarrow {\log }_{2}x=\frac{4}{{\log }_{2}x}$
$\Rightarrow {\left({\log }_{2}x\right)}^2=4$
$\Rightarrow {\log }_{2}x=\pm 2$
If ${\log }_{2}x=+2$
then $x=2^2=4$
and if ${\log }_{2}x=-2$,
then $x=2^{-2}=\frac{1}{4}$
$\therefore$ Solution set of the equation $=\left\{4,\frac{1}{4}\right\}$