If a^x = b^y = c^z and b² = ac, prove that :

y = $\dfrac{2xz}{x + z}$.

Given,

⇒ a^x = b^y = c^z = k (let)

⇒ a^x = k

⇒ a = $k^{\dfrac{1}{x}}$ ……..(1)

⇒ b^y = k

⇒ b = $k^{\dfrac{1}{y}}$ ……..(2)

⇒ c^z = k

⇒ c = $k^{\dfrac{1}{z}}$ ……..(3)

Given,

⇒ b² = ac

Substituting values of a, b and c in above equation, we get :

$\Rightarrow (k^{\dfrac{1}{y}})^2 = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{2}{y}} = k^{\dfrac{1}{x} + \dfrac{1}{z}} \\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z} \\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{z + x}{xz} \\[1em] \Rightarrow y = \dfrac{2xz}{x + z}.$

Hence, proved that $y = \dfrac{2xz}{x + z}.$