Sets, Relation and Function
116954
If \(\log 2+\frac{1}{2} \log a+\frac{1}{2} \log b=\log (a+b)\), then
1 \(\mathrm{a}=\mathrm{b}\)
2 \(\mathrm{a}=-\mathrm{b}\)
3 \(a=2, b=0\)
4 \(\mathrm{a}=10, \mathrm{~b}=1\)
Explanation:
A Given,
\(\log 2+\frac{1}{2} \log a+\frac{1}{2} \log b=\log (a+b)\)
or
\(\log 2+\log \sqrt{a}+\log \sqrt{b}=\log (a+b)\)
or \(\quad \log 2 \cdot \sqrt{a b}=\log a+b\)
\(\therefore \quad \mathrm{a}+\mathrm{b}=2 \cdot \sqrt{\mathrm{ab}}\)
or \(\quad(\sqrt{a}-\sqrt{b})^2=0\)
So, \(\quad \mathrm{a}=\mathrm{b}\)