Question

pls answer it fast....✌️❤️❤️❤️❤️❤️❤️❤️

Answer 1

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Answer 2

$Given \: \: Question \: \: Is \: \\ \\ a {}^{x} = b {}^{y} = c {}^{z} \: \: \: \: \: \: \: \: \: b {}^{2} = ac \\ \\ Answer \: \\ \\ a {}^{x} = b {}^{y} = c {}^{z} = k \: \: \: \: let \\ \\ \\ a {}^{x} = k \: \: \: \: \: \: b {}^{y} = k \: \: \: \: \: and \: \: \: \: c {}^{z} = k \\ \\ a = k {}^{ \frac{1}{x} } \: \: ...Equation \: \: \: i \\ \\ b = k {}^{ \frac{1}{y} } \: \: ...Equation \: \: \: \: ii \\ \\ c = k {}^{ \frac{1}{z} } \: \: ... \: \: Equation \: \: \: iii \\ \\ b {}^{2} = ac \: \: \: \: \: (Given) \\ \\ k {}^{ \frac{2}{y} } = k {}^{ \frac{1}{x} } \: \: . \: \: \: k {}^{ \frac{1}{z} } \\ \\ k {}^{ \frac{2}{y} } = k {}^{( \frac{1}{x} + \frac{1}{z} )} \\ \\ now \: compare \: powers \: of \: k \: \: we \: have \\ \\ \\ \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \\ \\ \frac{2}{y} = \frac{(z + x)}{xz} \\ \\ \frac{y}{2} = \frac{xz}{(x + z)} \\ \\ y = \frac{2zx}{(x + z)} \: \: \: \: hence \: \: \:proved \\ \\ \\ Note \: \\ \\ if \: \: t {}^{n} = v \: \\ then \: \: \: \: \: \: \: t = v {}^{ \frac{1}{n} }$