Question

If $2^x=4^y=8^z$ and $(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z})=\frac{24}{7}$, then find the value of z.

• A. $z=\frac{7}{48}$
• B. $z=\frac{7}{42}$
• C. $z=\frac{5}{48}$
• D. $z=\frac{9}{48}$

Answer 1

The correct option is A $z=\frac{7}{48}$

$2^x=4^y=8^z\Rightarrow 2^x=(2^2{)}^y=(2^3{)}^z$

$\Rightarrow 2^x=2^{2y}=2^{3z}\Rightarrow x=2y=3z$

$\therefore \frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{1}{2\times 3z}+\frac{1}{4\times \frac{3z}{2}}+\frac{1}{6z}$

$=\frac{1}{6z}+\frac{1}{6z}+\frac{1}{6z}=\frac{24}{7}$

$\Rightarrow \frac{3}{6z}=\frac{24}{7}$

$\Rightarrow \frac{1}{2z}=\frac{24}{7}\Rightarrow z=\frac{7}{2\times 24}\Rightarrow z=\frac{7}{48}$