Question 47

If $$\log \frac{a^{2}}{b}+\log \frac{b}{a^{2}}=\log(a+b)$$, then

Solution

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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 . 

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