If ${a^2} + {b^2} = 7ab$, prove that 2log(a+b)=log9+loga+logb

Hint-Express the LHS in the form of mloga=$\log {a^m}$ and solve it.

We have in the LHS, 2log(a+b)
So, making use of the formula mloga=$\log {a^m}$, we can write
LHS=2log(a+b)=$\log {(a + b)^2}$
Using the formula ${(a + b)^2} = {a^2} + 2ab + {b^2}$ ,
We can write
 $ \Rightarrow 2\log (a + b) = \log [{a^2} + {b^2} + 2ab]$
 In the question, it is already been given to us that
${a^2} + {b^2} = 7ab$,
Making use of this , we can write the equation as
$2\log (a + b) = \log [7ab + 2ab] = \log (9ab)$
Now , we also know the result which says
log(abc)=loga+logb+logc
So, we can write log(9ab)=log9+loga+logb=RHS
So, we have proved the result that 2log(a+b)=log9+loga+logb

Note: In accordance with the RHS which has to be proved, make use of the appropriate logarithmic formula and solve the problem.