alleryIf $3^x = 4^y = 12^z$ with $x,y,z$ nonzero, then $\dfrac{1}{z}$ equals
A$\dfrac{1}{x} + \dfrac{1}{y}$
B$x + y$
C$\dfrac{1}{x} - \dfrac{1}{y}$
D$xy$
Answer & Solution
Correct answer: A. $\dfrac{1}{x} + \dfrac{1}{y}$
Let $3^x=4^y=12^z=k$.
Then $3=k^{1/x}$, $4=k^{1/y}$, $12=k^{1/z}$.
Since $12 = 3\cdot4$: $k^{1/z}=k^{1/x}\cdot k^{1/y}=k^{1/x+1/y}$, so $\dfrac1z=\dfrac1x+\dfrac1y$.
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