If  $\log 2+\frac{1}{2}\log a+\frac{1}{2}\log b=\log (a+b)$ then

$\log 2+\frac{1}{2}\log a+\frac{1}{2}\log b=\log 2\sqrt{ab}$

$\Rightarrow \log 2\sqrt{ab}=\log (a+b)$

$\Rightarrow 2\sqrt{ab}=a+b$

Squaring on both sides

${\left(2\sqrt{ab}\right)}^2={(a+b)}^2$

$\Rightarrow 4ab=a^2+b^2+2ab$

$\Rightarrow a^2+b^2+2ab-4ab=0$

$\Rightarrow a^2+b^2-2ab=0$

$\begin{array}{l}\Rightarrow {(a-b)}^2=0\\ \Rightarrow a-b=0\\ \therefore a=b\end{array}$