Asked by Cindy
Given that x²+y²=5xy show that 2 log (x+y/√y)=log x+log y
Answers
Answered by
mathhelper
x^2 + y^2 = 5xy
x^2 + y^2 + 2xy = 7xy , I added 2xy to both sides
(x + y)^2 = 7xy
take log of both sides
log [(x+y)^2] = log(7xy)
log [(x+y)^2] = log7 + logx + logy
log [(x+y)^2] - log7 = logx + logy <---- I got the right side
2log(x+y) - log (√7)^2 = logx + logy
2log(x+y) - 2log (√7) = logx + logy
2 (log(x+y) - log (√7) ) = logx + logy
2log ( (x+y)/√7 ) = log x + log y <----- I have √7 instead of √y
I think you have a typo, check your original question.
x^2 + y^2 + 2xy = 7xy , I added 2xy to both sides
(x + y)^2 = 7xy
take log of both sides
log [(x+y)^2] = log(7xy)
log [(x+y)^2] = log7 + logx + logy
log [(x+y)^2] - log7 = logx + logy <---- I got the right side
2log(x+y) - log (√7)^2 = logx + logy
2log(x+y) - 2log (√7) = logx + logy
2 (log(x+y) - log (√7) ) = logx + logy
2log ( (x+y)/√7 ) = log x + log y <----- I have √7 instead of √y
I think you have a typo, check your original question.
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