Question

If ${\log }_4\left(2\right)+{\log }_4\left(4\right)+{\log }_4\left(x\right)+{\log }_4\left(16\right)=6$, then $x=$

• A. $64$
• B. $32$
• C. $8$
• D. $4$

Answer 1

The correct option is B

$32$

Explanation for the correct option:

Given: ${\log }_4\left(2\right)+{\log }_4\left(4\right)+{\log }_4\left(x\right)+{\log }_4\left(16\right)=6$

$$\begin{gathered} \Rightarrow {\log }_4\left(8\right)+{\log }_4\left(16x\right)=6\left[\because log\left(ab\right)=log\left(a\right)+log\left(b\right)\right] \\ \Rightarrow {\log }_4\left(8\times 16x\right)=6\mathbf{}\left[\because log\left(ab\right)=log\left(a\right)+log\left(b\right)\right] \\ \Rightarrow {\log }_4\left(128x\right)=6 \\ \Rightarrow 128x=4^6\left[\because lo{g}_a\left(b\right)=c\Rightarrow b=a^c\right] \\ \Rightarrow x=\frac{4096}{128} \\ \Rightarrow x=32 \end{gathered}$$

Hence option (B) i.e. $32$ is correct.