Solve the equations:
$3^{x} 5^{y} = 75$
$3^{y} 5^{x} = 45$

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Solution

Dividing both the equations, we get
$3^{x - y} \times 5^{y - x} = \frac{75}{45}$

${(\frac{3}{5})}^{x - y} = {(\frac{3}{5})}^{- 1}$

Comparing the powers as bases are equal,

$⟹ x - y = - 1$ .............(1)

Again mulitiplying the equation we get

$(3^{x} \times 3^{y}) (3^{x} \times 3^{y}) = 75 \times 45$

$15^{x + m y} = 15^{3}$

$⟹ x + y = 3$ ............(2)

solving (1) and (2) we get ,

$x = 1, y = 2$

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