Question

The $\log \left(\frac{a+b}{2}\right)=\frac{1}{2}(\log a+\log b)$, then show that $a=b$.

Answer 1

$\log \left(\frac{a+b}{2}\right)=\frac{1}{2}(\log a+\log b)$

$\log \left(\frac{a+b}{2}\right)=\frac{1}{2}\left(\log ab\right)$

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

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

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

$a+b-2\sqrt{ab}=0$

$(\sqrt{a}-\sqrt{b}{)}^2=0$

$\sqrt{a}-\sqrt{b}=0$

$\sqrt{a}=\sqrt{b}$

$\therefore a=b$