If (3)^x = (5)^y = (75)^z, show that $z = \dfrac{xy}{2x + y}$

Let (3)^x = (5)^y = (75)^z = k

∴ (3)^x = k

⇒ 3 = $k^{\dfrac{1}{x}}$ ……(i)

∴ (5)^y = k

⇒ 5 = $k^{\dfrac{1}{y}}$ …….(ii)

∴ (75)^z = k

⇒ 75 = $k^{\dfrac{1}{z}}$

⇒ 3.5² = $k^{\dfrac{1}{z}}$

Substituting value of x and y from (i) and (ii) in above equation we get,

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

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