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Examveda

If $${\log _x}\left( {\frac{9}{{16}}} \right) = - \frac{1}{2},$$    then x is equal to:

A. $$ - \frac{3}{4}$$

B. $$\frac{3}{4}$$

C. $$\frac{{81}}{{256}}$$

D. $$\frac{{256}}{{81}}$$

Answer: Option D

Solution(By Examveda Team)

$$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr} $$

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