Question

If ${\log }_{2}x+{\log }_{8}x+{\log }_{64}x=3$, then the value of $x$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

${\log }_{2}x+{\log }_{8}x+{\log }_{64}x=3$
$\Rightarrow {\log }_{2}x+{\log }_{2^3}x+{\log }_{2^6}x=3$
$\Rightarrow {\log }_{2}x+\frac{1}{3}{\log }_{2}x+\frac{1}{6}{\log }_{2}x=3\Rightarrow \frac{9}{6}{\log }_{2}x=3\Rightarrow {\log }_{2}x=2\Rightarrow x=4$