A) \[\left\{ 4,\frac{1}{4} \right\}\]
B) \[\left\{ 2,\frac{1}{2} \right\}\]
C) \[\{1,2\}\]
D) \[\left\{ 8,\frac{1}{8} \right\}\]
Correct Answer: A
Solution :
\[\therefore \] \[{{\left[ 4\left( -\frac{1}{3}+\frac{1}{9}-\frac{1}{27} \right)+... \right]}^{{{\log }_{2}}x}}\] \[={{\left[ 54\left( 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+... \right) \right]}^{{{\log }_{x}}2}}\] \[\Rightarrow \] \[{{\left[ 4\left( \frac{1}{1+\frac{1}{3}} \right) \right]}^{{{\log }_{2}}x}}={{\left[ 54\left( \frac{1}{1-\frac{1}{3}} \right) \right]}^{{{\log }_{x}}2}}\] \[\Rightarrow \] \[{{\left[ 4\left( \frac{3}{4} \right) \right]}^{{{\log }_{2}}x}}={{\left[ 54\times \frac{3}{2} \right]}^{{{\log }_{x}}2}}\] \[\Rightarrow \] \[{{3}^{{{\log }_{2}}x}}={{(81)}^{{{\log }_{x}}2}}\] \[\Rightarrow \] \[{{3}^{{{\log }_{2}}x}}={{3}^{4{{\log }_{x}}2}}\] \[\Rightarrow \] \[{{\log }_{2}}x=4{{\log }_{x}}2\] \[\Rightarrow \] \[{{\log }_{2}}x=\frac{4}{{{\log }_{2}}x}\] \[\Rightarrow \] \[{{({{\log }_{2}}x)}^{2}}=4\] \[\Rightarrow \] \[{{\log }_{2}}x=\pm \,2\] 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 \[=\left\{ 4,\frac{1}{4} \right\}.\]
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