Question

\[ 2^x = 3^y = 6^{-z} \]

Solution

Let \(2^x = 3^y = 6^{-z} = k\) \[ x = \log_2 k,\quad y = \log_3 k,\quad -z = \log_6 k \] \[ \frac{1}{x} = \log_k 2,\quad \frac{1}{y} = \log_k 3,\quad \frac{1}{z} = -\log_k 6 \] \[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_k 2 + \log_k 3 – \log_k 6 \] \[ = \log_k (2 \cdot 3) – \log_k 6 \] \[ = \log_k 6 – \log_k 6 \] \[ = 0 \]

Answer

\[ \boxed{0} \]

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