Challenge yourself!

Method 1: we have that $\large 2^{x} = 6^{-z} \Longrightarrow 2 = 6^{-z/x}$ and that $\large 3^{y} = 6^{-z} \Longrightarrow 3 = 6^{-z/y}$ .

Multiplying, we have that $\large 2*3 = 6^{-z/x}*6^{-z/y} \Longrightarrow 6^{1} = 6^{-z/x - z/y}$ .

Equating exponents gives us that $1 = -\dfrac{z}{x} - \dfrac{z}{y} \Longrightarrow \dfrac{1}{z} = -\dfrac{1}{x} - \dfrac{1}{y} \Longrightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \boxed{0}$ .

Method 2: Taking logs, we have that $2^{x} = 3^{y} \Longrightarrow x\ln(2) = y\ln(3) \Longrightarrow \dfrac{1}{y} = \dfrac{1}{x}*\dfrac{\ln(3)}{\ln(2)}$ ,

and that $2^{x} = 6^{-z} \Longrightarrow x\ln(2) = -z\ln(6) \Longrightarrow \dfrac{1}{z} = -\dfrac{1}{x}*\dfrac{\ln(6)}{\ln(2)}$ .

Thus $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{x}\left(1 + \dfrac{\ln(3)}{\ln(2)} - \dfrac{\ln(6)}{\ln(2)}\right) = \dfrac{1}{x}\dfrac{\ln(2) + \ln(3) - \ln(6)}{\ln(2)} = \boxed{0}$

since $\ln(6) = \ln(2*3) = \ln(2) + \ln(3)$ .