Question

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

Answer 1

Given:$2^x=4^y=8^z\Rightarrow 2^x=2^{2y}=2^{3z}\Rightarrow x=2y=3z=k-\left(i\right)$

$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{24}{7}-\left(ii\right)$ (given)

Substitute $\left(i\right)$ in $\left(ii\right)$ ; $\frac{1}{2k}+\frac{1}{2k}+\frac{1}{2k}=\frac{24}{7}\Rightarrow \frac{3}{2k}=\frac{24}{7}\Rightarrow k=\frac{7}{16}$

$\therefore 3z=k\Rightarrow z=\frac{k}{3}=\frac{7}{48}$