Question

If $a^2+b^2=7ab$. Prove that $\log \left(\frac{1}{3}(a+b)\right)=\frac{1}{2}[\log a+\log b]$

Answer 1

Given that $a^2+b^2=7ab$

Adding $2ab$ on both sides, we get

$a^2+b^2+2ab=9ab$

$\Rightarrow (a+b{)}^2=9ab$

$\Rightarrow a+b=\sqrt{9ab}$

$\Rightarrow a+b=3(ab{)}^{\frac{1}{2}}$

$\Rightarrow \frac{1}{3}(a+b)=(ab{)}^{\frac{1}{2}}$

Applying $\log$ on both sides we get

$\Rightarrow \log \left(\frac{1}{3}\right(a+b\left)\right)=\log (ab{)}^{\frac{1}{2}}$

$\Rightarrow \log \left(\frac{1}{3}\right(a+b\left)\right)=\frac{1}{2}[\log a+\log b]$

$[\because \log m^n=n\log mand\log mn=\log m+\log n]$