Question

If $2^x=3^y=6^{-z}$, show that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

Answer 1

Let $2^x=3^y=6^{-z}=k$

$\Rightarrow 2^x=k,3^y=k,6^{-z}=k$

$\Rightarrow 2=k^{\frac{1}{x}},3=k^{\frac{1}{y}},{(2\times 3)}^{-z}=k$

$\Rightarrow 2=k^{\frac{1}{x}},3=k^{\frac{1}{y}},2\times 3=k^{-\frac{1}{z}}$

$\Rightarrow 2=k^{\frac{1}{x}},3=k^{\frac{1}{y}},k^{\frac{1}{x}}{k}^{\frac{1}{y}}=k^{-\frac{1}{z}}$

$\Rightarrow k^{\frac{1}{x}+\frac{1}{y}}=k^{-\frac{1}{z}}$

$\Rightarrow \frac{1}{x}+\frac{1}{y}=-\frac{1}{z}$

$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

Hence proved.