Solution
We can simplify the LHS to $$\log\ \left(\frac{a^2}{b}\times\ \frac{b}{a^2}\right)\ =\log\ 1$$
Now log 1 = log(a+b) Hence a+b=1 .
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We can simplify the LHS to $$\log\ \left(\frac{a^2}{b}\times\ \frac{b}{a^2}\right)\ =\log\ 1$$
Now log 1 = log(a+b) Hence a+b=1 .
