Asked by Anonymous
Given that x squared + y squared = 5xy,show that; (a) 2log(x+y\√7)=log x + log y. (b). Log ( x- y\√3) = 1\2( log x + log y)
Answers
Answered by
Reiny
x^2 + y^2 = 5xy
complet the square:
x^2 + 2xy + y^2 = 7xy
(x+y)^2 = 7xy
take log of both sides
2log(x+y) = log 7 + logx + logy
2log(x+y) - log7 = logx + logy
2log(x+y) - 2log 7^(1/2) = logx + logy
2[ log(x+y) - log √7] = logx + logy
2 log( (x+y)/√7 ) = logx + logx
as required
to prove the second one, I suggest:
x^2 - 2xy + y^2 = 3xy
and follow the same steps as above
complet the square:
x^2 + 2xy + y^2 = 7xy
(x+y)^2 = 7xy
take log of both sides
2log(x+y) = log 7 + logx + logy
2log(x+y) - log7 = logx + logy
2log(x+y) - 2log 7^(1/2) = logx + logy
2[ log(x+y) - log √7] = logx + logy
2 log( (x+y)/√7 ) = logx + logx
as required
to prove the second one, I suggest:
x^2 - 2xy + y^2 = 3xy
and follow the same steps as above
Answered by
ayanfe
absolutely wrong
Answered by
ayanfe
correctl
Answered by
Michael omoloye
Thanks you bro
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