Question

If $2^x=3^y={12}^z$, prove that $-\frac{2}{x}+\frac{1}{y}+\frac{1}{z}=0$

Answer 1

$2^x=3^y={12}^z=K$

$2=K^{\frac{1}{x}},3=K^{\frac{1}{y}},12=K^{\frac{1}{z}}$

$2^x=K$ & $3^y=K$ & ${12}^z=K$

Apply log on both sides.

$xlog2=logkylog3=logkzlog12=logk$

$\frac{1}{x}=\frac{lo{g}_2}{lo{g}_k}\frac{1}{y}=\frac{lo{g}_3}{lo{g}_k}\frac{1}{z}=\frac{lo{g}_{1}2}{lo{g}_k}$

$\frac{1}{x}=lo{g}_k^2\frac{1}{y}=lo{g}_k^3\frac{1}{z}=lo{g}_k^{12}$

$\frac{1}{z}=lo{g}_k(2^2\times 3)=log{2}_k^2+lo{g}_k^3=2lo{g}_k^2+lo{g}_k^3$

$\frac{1}{z}=\frac{2}{x}+\frac{1}{y}$ Hence proved.