Question

Let $a^4+b^4=7a^2{b}^2$ then prove that $\log (a^2+b^2)=\log a+\log b+\log 3$.

Answer 1

Given,

$a^4+b^9=7a^2{b}^2$

or $(a^2+b^2{)}^2-2a^2{b}^2=7a^2{b}^2$

or, $(a^2+b^2{)}^2=9a^2{b}^2$

Now, taking logarithm both sides w.r.t. to the base 'e' we get,

$2log(a^2+b^2)=log9+log{a}^2+log{b}^2$

or, $2log(a^2+b^2)=2log3+2loga+2logb$

or, $log(a^2+b^2)=log3+loga+logb$.