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Question 133
If $$\frac{1}{2} \log x +Â \frac{1}{2} \log y + \log 2 = \log(x + y)$$, then ..............
Solution
mlog(a) + nlog(b) = log($$a^{\frac{1}{m}}\cdot b^{\frac{1}{n}}$$)
Therefore the given LHS reduces to log(2$$\sqrt{\ xy}$$) which is equal to log(x+y)
Remove log on both sides
2$$\sqrt{\ xy}$$=x+y
$$\left(\sqrt{\ x}\right)^{2\ }+\ \left(\sqrt{\ y}\right)^2\ -2\sqrt{\ x\cdot y}=0$$
$$\left(\sqrt{\ x}-\sqrt{\ y}\right)^2=0$$
=> x=y
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