Alg9.4.26

If $a^2+b^2=7ab\,$ then show that $2\log(a+b)=2\log 3+\log a+\log b\,$

Given that

$a^2+b^2=7ab\,$

Adding 2ab on both sides,we get

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

$(a+b)^2=9ab\,$

Applying logs on both sides,we get

$\log (a+b)^2=\log 9ab\,$

Simplifying

$2\log(a+b)=\log 9+\log a+\log b\,$

$2\log(a+b)=2\log 3+\log a+\log b\,$

This is equal to the equation to be proved.