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

Given:

\( 3^{x}=5^{y}=(75)^{z} \)

To do: 

We have to show that \( z=\frac{x y}{2 x+y} \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

Let $3^{x}=5^{y}=75^{z}=k$

This implies,

$3=k^{\frac{1}{x}}, 5=k^{\frac{1}{y}}$ and $75=k^{\frac{1}{z}}$

$75=3\times25=3\times5^2$

$\Rightarrow 3 \times 5^{2}=k^{\frac{1}{z}}$

$\Rightarrow k^{\frac{1}{x}} \times k^{\frac{2}{y}}=k^{\frac{1}{z}}$

$\Rightarrow k^{\frac{1}{x}+\frac{2}{y}}=k^{\frac{1}{z}}$

Comparing both sides, we get,

$\Rightarrow \frac{1}{x}+\frac{2}{y}=\frac{1}{z}$

$\Rightarrow \frac{y+2 x}{x y}=\frac{1}{z}$

$\Rightarrow z=\frac{x y}{2 x+y}$

Hence proved.