Question

If $log\frac{x}{y}+log\frac{y}{x}=log(x+y)$, then which of the following equations is true?

• A. $x+y=1$
• B. $x-y=1$
• C. $x+y=0$
• D. $x^2+y^2=1$

Answer 1

The correct option is A

$x+y=1$

Simplifying LHS, $\log \frac{x}{y}+log\frac{y}{x}=log(\frac{x}{y}\times \frac{y}{x})=log1$

So,

$\Rightarrow log(x+y)=log1$

$\Rightarrow x+y=1$