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Thursday, November 11, 2021

2021/093) Given 2^x = 3^y = 6^{-z} evaluate \frac{1}{x} +\frac{1}{y} + \frac{1}{z}

we are given 2^x = 3^y = 6^{-z}

so 2 = 6^{\frac{-z}{x}}

and 3 = 6^\frac{-z}{y}

so 2 * 3 = 6^{(-\frac{z}{x} - \frac{z}{y})}

or 6^1 = 6^{(-\frac{-z}{x} - \frac{z}{y})}


or 1 = -\frac{z}{x} - \frac{z}{y}

or  \frac{z}{x} +\frac{z}{y} + 1 = 0

or \frac{1}{x} +\frac{1}{y} + \frac{1}{z}= 0

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