Question

After evaluating ${\log }_{3}x+{\log }_{3}x={\log }_{\sqrt{3}}x+{\log }_3\sqrt{x}+\frac{1}{2}$, value of $x$ will be

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{1}{3}$

${\log }_{3}x+{\log }_{3}x={\log }_{\sqrt{3}}x+{\log }_3\sqrt{x}+\frac{1}{2}$

$\Rightarrow 2{\log }_{3}x=2{\log }_{3}x+\frac{1}{2}{\log }_{3}x+\frac{1}{2}\dots ({\log }_{3^a}x=\frac{1}{a}{\log }_{3}x)$

$\Rightarrow \frac{1}{2}{\log }_{3}x=-\frac{1}{2}$

$\Rightarrow {\log }_{3}x=-1$

$\Rightarrow x=3^{-1}$

$\Rightarrow x=\frac{1}{3}$