Question

If $2^x=4^y=8^{z}andxyz=288$, then $\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}$ equal

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

Answer 1

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

$2^x=(2^2{)}^y=(2^3{)}^z\Rightarrow x=2y=3z$
Given $xyz=288$ $\Rightarrow x\times \frac{x}{2}\times \frac{x}{3}=288$
$\Rightarrow x^3=6\times 288\Rightarrow x^3=1728\Rightarrow x=\sqrt[3]{31728}=12$
$\therefore y=\frac{12}{2}=6$ and $z=\frac{12}{3}=4$
$\therefore \frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}=\frac{1}{24}+\frac{1}{24}+\frac{1}{32}=\frac{4+4+3}{96}=\frac{11}{96}$