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 (y-9)(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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