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

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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.

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.

Last updated date: 26th Sep 2023

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