Posted by Anonymous on Friday, July 26, 2013 at 10:11pm.
solve 2y=7sqrt(y)3

math  Reiny, Friday, July 26, 2013 at 10:17pm
2y=7sqrt(y)3
2y+3 = √y
square both sides
4y^2 + 12y + 9 = y
4y^2 + 11y + 9 = 0
y = (11 ±√23)/8
I am assuming you are dealing with real numbers only
since y is a complex number, and we cannot take a square root of a complex number, your equation has no real solution.

math  Anonymous, Friday, July 26, 2013 at 10:24pm
Using the substitution y=e^x,solve 2e^x=7√e^x 3

math  Reiny, Friday, July 26, 2013 at 10:42pm
for some reason , I dropped the 7 on the right side
let's start again
square both sides
4y^2 + 12y + 9 = 49y
4y^2  37y + 9 = 0
y = (37 ± √1225)/8
y = 9 or y = 0.25
(it would have factored to (y9)(4y  1)=0 )
now now
e^x = 9
take ln of both sides
ln (e^x) = ln9
x = ln9
or
e^x = .25 or 1/4
take ln again
x = ln (1/4) = ln1  ln4 = ln4
x = ln9 or x = ln4
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