Question

If $2^x=4^y=8^z$ and $xyz=288$ then $\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}$ is equal to

• A. $\frac{11}{8}$
• B. $\frac{11}{24}$
• C. $\frac{11}{48}$
• D. $\frac{11}{96}$

Answer 1

The correct option is D $\frac{11}{96}$

If $2^x=4^y=8^z$ i.e., $2^x=2^{2y}=2^{3z}$
Since base are equal hence,
$x=2y=3z$ .........(1)
Also given that $xyz=288$ .......(2)
From (1), we have
$y=\frac{x}{2}$ and $z=\frac{x}{3}$
Substituting $y=\frac{x}{2}$ and $z=\frac{x}{3}$ in (2), we get
$\therefore$ $x=12$
Re substituting $x=12$ in $y=\frac{x}{2}$ and $z=\frac{x}{3}$, we get
$y=6$ and $z=4$
$\therefore$$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}$ is equal to $\frac{11}{96}$