If \(2^{x} = 3^{y} = 6^{-z}\), then \((\frac{1}{x} + \frac{1}{y} + \frac{1}{z})\) is equal to
Question and options
If , then is equal to
- 0
- 1
- 2
- 4
Answer and explanation
Correct answer: A - 0
Let 2^x = 3^y = 6^-z = k. Then 1/x + 1/y + 1/z = (log2 + log3 - log6)/logk = 0.
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