Question

If $a^2+b^2=14a^2{b}^2$ then prove that $\log (a+b)=2\log 2+\log a+\log b$.

Answer 1

Given $a^2+b^2=14a^2{b}^2$

or, $(a+b{)}^2=16a^2{b}^2$

Now taking logarithm both sides with respect to the base $e$ we get,

$2\log (a+b)=\log 16+\log a^2+\log b^2$

or, $2\log (a+b)=4\log 2+2\log a+2\log b$

or, $\log (a+b)=2\log 2+\log a+\log b$>