Question

If ${\log }_{4}2+{\log }_{4}4+{\log }_{4}x+{\log }_{4}16=6$, then value of $x$ is

• A. $64$
• B. $4$
• C. $8$
• D. $32$

Answer 1

The correct option is D $32$

We have ${\log }_{4}2+{\log }_{4}4+{\log }_{4}x+{\log }_{4}16=6$

$\Rightarrow {\log }_4(2\times 4\times x\times 16)=6$
$\Rightarrow 128x=4^6$

$\Rightarrow 4^3\times 2x=4^6$

$\Rightarrow x=\frac{4^3}{2}$

$\Rightarrow x=32$