Question

2^x=3^y=6^-z=k.prove that 1/x+1/y=1/z?

Answer 1

$Given \: \: Question \: \: Is \: \\ \\ 2 {}^{x} = 3 {}^{y} = 6 {}^{ - z} \\ \\ Answer \: \\ \\ let \: \: \: \: 2 {}^{x} = 3 {}^{y} = 6 {}^{ - z} \: \: = k \\ \\ 2 {}^{x} = k \: \: \: \: 3 {}^{y} = k \: \: \: \: \: and \: \: \: 6 {}^{ - z} = k \\ \\ 2 = k {}^{ \frac{1}{x} } \: \: ...Equation \: \: i \: \\ \\ 3 = k {}^{ \frac{1}{y} } \: \: ...Equation \: \: \: ii \\ \\ 6 = k {}^{ \frac{ - 1}{z} } \: \: \: ... \: Equation \: \: \: iii \\ \\ \\ 6 = k { }^{ \frac{ - 1}{z} } \\ \\ 2 \times 3 = k {}^{ \frac{ - 1}{z} } \: \: \: \: \: from \: Equation \: i \: and \:ii \\ \\ k {}^{ \frac{1}{x} } \times k {}^{ \frac{1}{y} } = k {}^{ \frac{ - 1}{z} } \\ \\ k {}^{( \frac{1}{x} + \frac{1}{y}) } = k {}^{ \frac{ - 1}{z} } \\ \\ Now \: \: compare \: powers \: of \: k \: we \: have \\ \\ \frac{1}{x} + \frac{1}{y} = \frac{ - 1}{z} \: \: \: Hence \: proved \: \\ \\ Note \: \\ \\ if \: \: \: x {}^{n} = k \\ \\ then \: \: \: x = k {}^{ \frac{1}{n} }$