Question
If $${\log _x}\left( {\frac{9}{{16}}} \right) = - \frac{1}{2},$$ then x is equal to:
Answer: Option D
$$\eqalign{
& {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr
& \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr
& \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr
& \Rightarrow \sqrt x = {{16} \over 9} \cr
& \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr
& \Rightarrow x = {{256} \over {81}} \cr} $$
Was this answer helpful ?
$$\eqalign{
& {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr
& \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr
& \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr
& \Rightarrow \sqrt x = {{16} \over 9} \cr
& \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr
& \Rightarrow x = {{256} \over {81}} \cr} $$
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution