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)

$\therefore \left[4\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\ldots\right)\right]^{\log _{2} x}$
$=\left[54\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\right)\right]^{\log _{x} 2}$
$\Rightarrow \left[4\left(\frac{1}{1+\frac{1}{3}}\right)\right]^{\log _{2} x}=\left[44\left(\frac{1}{1-\frac{1}{3}}\right)\right]^{\log _{x} 2}$
$\Rightarrow \left[4\left(\frac{3}{4}\right)\right]^{\log _{2} x}=\left[54 \times \frac{3}{2}\right]^{\log _{x} 2}$
$\Rightarrow 3^{\log _{2} x}=(81)^{\log _{x} 2}$
$\Rightarrow 3^{\log _{2} x}=3^{4 \log _{x} 2}$
$\Rightarrow \log _{2} x=4 \log _{x} 2$
$\Rightarrow \log _{2} x=\frac{4}{\log _{2} x}$
$\Rightarrow \left(\log _{2} x\right)^{2}=4$
$\Rightarrow \log _{2} x=\pm 2$
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 $=\left\{4, \frac{1}{4}\right\}$