Question

anyone do this sum please..​

Answer 1

$here \: is \: your \: answer \\ \\ 3 {}^{x} = 5 {}^{x} = (75) {}^{z} = k \: \: say \: \: \\ where \: k \: belongs \: to \: the \: set \: of \: real \: numbers \\ \\ 3 {}^{x} = k \: ...(01) \\ \\ 5 {}^{y} = k \: \: ...(02) \\ \\ (75) {}^{z} = k \: \: ...(03) \\ \\ from \: \: third \\ \\ 5 {}^{2} \times 3 = k \\ \\ k {}^{ \frac{2}{y} } \times k {}^{ \frac{1}{x} } = k {}^{ \frac{1}{z} } \\ \\ k {}^{ \frac{2}{y} + \frac{1}{x} } = k {}^{ \frac{1}{z} } \\ now \: compare \: powers \: of \: k \: \\ \\ \frac{2}{y} + \frac{1}{x} = \frac{1}{z} \\ \\ z = \frac{xy}{2x + y}$