Question

If $a^2+b^2=3ab,$ show that $\log \left(\frac{a+b}{\sqrt{5}}\right)=\frac{1}{2}(\log a+\log b)$

Answer 1

We have,

$a^2+b^2=3ab$

On adding $2ab$ both side and we get,

$a^2+b^2+2ab=3ab+2ab$

$\Rightarrow {(a+b)}^2=5ab$

On taking $\log$ both side

$\frac{(a+b)}{\sqrt{5}}=\sqrt{ab}$

$\Rightarrow \log \frac{(a+b)}{\sqrt{5}}=\log \sqrt{ab}$

$\Rightarrow \log \frac{(a+b)}{\sqrt{5}}=\log \sqrt{a}+\log \sqrt{b}$

$\Rightarrow \log \frac{(a+b)}{\sqrt{5}}=\log {\left(a\right)}^{\frac{1}{2}}+\log {\left(b\right)}^{\frac{1}{2}}$

$\Rightarrow \log \frac{(a+b)}{\sqrt{5}}=\frac{1}{2}\mathrm{loga}+\frac{1}{2}\log b$

$\Rightarrow \log \frac{(a+b)}{\sqrt{5}}=\frac{1}{2}(\mathrm{loga}+\log b)$

Hence, this is the answer.