Question

If $4^x=8^y={16}^z$ then prove that $\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

Answer 1

Given,

$4^x=8^y={16}^z$

$2^{2x}=2^{3y}=2^{4z}$

By comparing

$2x=3y=4z$ .......(1)

$R.H.S=$ $\frac{1}{x}+\frac{1}{z}$

$=$ $\frac{1}{\frac{3y}{2}}+$$\frac{1}{\frac{3y}{4}}$ {using (1)}

$=$ $\frac{2}{3y}+\frac{4}{3y}$

$=$ $\frac{2+4}{3y}$

$=$ $\frac{6}{3y}$

$=$ $\frac{2}{y}$

$=$ $L.H.S$