Question

If $4^x=8^y={16}^z$, then prove that $\left(\frac{2}{y}\right)$=$(\frac{1}{x}+\frac{1}{z}).$

Answer 1

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

$\Rightarrow (2^2{)}^x=(2^3{)}^y=(2^4{)}^7$

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

$\Rightarrow log\left(2^{2x}\right)=log\left(2^{3y}\right)=log\left(2^{4z}\right)$

$\Rightarrow 2xlog2=3ylog2=4zlog2$

$\Rightarrow 2x=3y=4z$ $(\because log2\ne 0)$

$\Rightarrow x=\frac{3y}{2}$ and $z=\frac{3y}{4}$

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

$\therefore \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$