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