Question

if A^x = B^y = C^z and A^2 = BC then z equals to​

Answer 1

Given,

$A^x = B^y = C^z \: and \: A^2 = BC$

From the above,

$\displaystyle \: \longrightarrow \: A^x = B^y \\ \longrightarrow \: A^ \frac{x}{y} = B \\ \\ \\ \longrightarrow \: A ^x = C^z \\ \longrightarrow \: A ^ \frac{x}{z} = C$

Now,

$\rightarrow \: A^2 = BC \\ \rightarrow A^2 \: = {A}^{ \frac{x}{y} } \times {A}^{ \frac{x}{z} } \\ \rightarrow \: A^2 \: = {A}^{ \frac{x}{y} + \frac{x}{z} } \\ \rightarrow 2 = x( \frac{1}{y} + \frac{1}{z} ) \\ \rightarrow \frac{2}{x} = ( \frac{1}{y} + \frac{1}{z} ) \\ \rightarrow \frac{2}{x} - ( \frac{1}{y}) = ( \frac{1}{z} ) \\ \\ \rightarrow \frac{2y - x}{xy} = \frac{1}{z} \\ \rightarrow \boxed{ z = \frac{xy}{2y - x} }$

Therefore, $\boxed{ z = \frac{xy}{2y - x} }$