# If 2 x </msup> = 3 y </msup> = 6 &#x2

If ${2}^{x}={3}^{y}={6}^{-z}$ and $x,y,z\ne 0$ then prove that:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
I have tried starting with taking logartithms, but that gives just some more equations.
Any specific way to solve these type of problems?
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rioolpijpgp
${2}^{x}={3}^{y}={6}^{-z}=k$ say, then $2={k}^{\frac{1}{x}},3={k}^{\frac{1}{y}},6={k}^{\frac{-1}{z}}$ now can you go on?
then ${k}^{\frac{-1}{z}}=6=2×3={k}^{\frac{1}{x}}×{k}^{\frac{1}{y}}={k}^{\frac{1}{x}+\frac{1}{y}}$

arridsd9
${2}^{x}={3}^{y}={6}^{-z}=k$
so
$x={\mathrm{log}}_{2}k$
$y={\mathrm{log}}_{3}k$
$z=-{\mathrm{log}}_{6}k$
so
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\mathrm{log}}_{k}2+lo{g}_{k}3-{\mathrm{log}}_{k}6$
$={\mathrm{log}}_{k}\frac{2×3}{6}$
$=0$