Question

If x2 + y2 = 7xy, then prove that log(x+y)/3 = 1/2(log x +logy)
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Answer 1

Step-by-step explanation:

Given :-

x^2+y^2 = 7xy

To Prove :-

Prove that log(x+y)/3 = 1/2(log x +logy)

Solution:-

Given equation is x^2+y^2 = 7xy

On adding "2xy " both sides then

=>x^2+y^2 + 2xy = 7xy+2xy

LHS is in the form of a^2+b^2+2ab .

Where a = x and b=y

We know that (a+b)^2=a^2+2ab+b^2

=>(x+y)^2 = 9xy

On taking Logarithms both sides then

=>log(x+y)^2 = log(9xy)

We know that log a^m = m log a

=>2 log (x+y) = log(9xy)

=>2 log (x+y) = log(3^2xy)

We know that log ab = log a + log b

=>2 log (x+y) = log 3^2 + log x + log y

We know that log a^m = m log a

=>2 log (x+y) =2 log 3+ log x + log y

On dividing by 2 both sides then

=>[2 log (x+y) ]/2 =[2 log 3+ log x + log y]/2

=> log (x+y) = log 3+ (log x + log y)/2

=>log (x+y) -log 3 =( log x + log y)/2

We know that log a - log b = log (a/b)

=>log (x+y)/3 = (log x+ log y)/2

Therefore, log (x+y)/3 = 1/2 (log x+ log y)

Hence Proved.

Used formulae:-

• (a+b)^2=a^2+2ab+b^2

• log ab = log a + log b

• log a - log b = log (a/b)

• log a^m = m log a