Evaluate log base xy of 1/x+ log base xy of 1/y

$\log_{x y} (\frac{1}{x}) + \log_{x y} (\frac{1}{y})$

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{x y} (\frac{1}{x} \cdot \frac{1}{y})$

Multiply $\frac{1}{x}$ by $\frac{1}{y}$ .

$\log_{x y} (\frac{1}{x y})$

Rewrite $\frac{1}{x y}$ as $1 {(x y)}^{- 1}$ .

$\log_{x y} (1 {(x y)}^{- 1})$

Rewrite $\log_{x y} (1 {(x y)}^{- 1})$ as $\log_{x y} (1) + \log_{x y} ({(x y)}^{- 1})$ .

$\log_{x y} (1) + \log_{x y} ({(x y)}^{- 1})$

Use logarithm rules to move $- 1$ out of the exponent.

$\log_{x y} (1) - \log_{x y} (x y)$

Logarithm base $x y$ of $x y$ is $1$ .

$\log_{x y} (1) - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$\log_{x y} (1) - 1$

Logarithm base $x y$ of $1$ is $0$ .

$0 - 1$

Subtract $1$ from $0$ .

$- 1$