Question

If ${\log }_{2}x+{\log }_{4}x+{\log }_{16}x=\frac{21}{4}$, then the value of $x$ is

• A. $10$
• B. $9$
• C. $8$
• D. $7$

Answer 1

The correct option is C $8$

${\log }_{2}x+{\log }_{4}x+{\log }_{16}x=\frac{21}{4}$

${\log }_{2}x+{\log }_{2^2}x+{\log }_{2^4}x=\frac{21}{4}$

${\log }_{2}x+\frac{1}{2}{\log }_{2}x+\frac{1}{4}{\log }_{2}x=\frac{21}{4}$

$\frac{7}{4}{\log }_{2}x=\frac{21}{4}$

${\log }_{2}x=3$

$\Rightarrow x=2^3=8$