Question

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

Answer 1

Given that,

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

Then prove that

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

Given $a^2+b^2=3ab...................\left(1\right)$

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

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

$a^2+b^2+2ab=5ab$

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

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

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

Taking $\log$ both side and we get,

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

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

Since, $(\log x^n=n\log x)$

Therefore,

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

Hence proved.