If $log \dfrac{a}{b}+log \dfrac{b}{a}$=log(a+b),then
a + b = 1
a - b = 1
a = b
$a^{2}-b^{2}=1$
Explanation:
$log \dfrac{a}{b}+log \dfrac{b}{a}$=log(a+b)
=>log(a+b)=$log(\dfrac{a}{b} \times \dfrac{b}{a})$=log 1.
so,a+b=1
$log \dfrac{a}{b}+log \dfrac{b}{a}$=log(a+b)
=>log(a+b)=$log(\dfrac{a}{b} \times \dfrac{b}{a})$=log 1.
so,a+b=1