Question

Prove the following identities :
If $a^2+b^2=7ab$ , prove that $\log \frac{1}{3}(a+b)=\frac{1}{2}[\log a+\log b]$

Answer 1

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

Adding $2ab$ on both sides we get,

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

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

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

Now applying $\log$ on both sides we get,

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

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

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