Question 133

If $$\frac{1}{2} \log x + \frac{1}{2} \log y + \log 2 = \log(x + y)$$, 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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