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
No comments:
Post a Comment