SOLUTION: how to solve this equation: 3(ln2x)^3-4(ln2x)^2-5ln2x+2=0

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Question 487617: how to solve this equation:

3(ln2x)^3-4(ln2x)^2-5ln2x+2=0

Found 2 solutions by CubeyThePenguin, ikleyn:
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
Let ln(x) = y.

3y^3 - 4y^2 - 5y + 2 = 0
(y - 2)(3y - 1)(y + 1) = 0

y = 2, y = 1/3, or y = -1

ln(x) cannot be negative, eliminating the last case.

ln(x) = 2 ----> x = e^2
ln(x) = 1/3 ----> x = 1/(e^3)

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

            @ThePenguin solved the problem incorrectly, by incorrectly presenting the second solution and by loosing the third solution.

            I came to bring a correct version.


Let ln(x) = y.


3y^3 - 4y^2 - 5y + 2 = 0
(y - 2)(3y - 1)(y + 1) = 0


y = 2, y = 1/3, or y = -1



        Consider all three cases separately. 



1)  ln(x) = 2    ---->  x = e^2.


2)  ln(x) = 1/3 ---->  x = e^(1/3) = root%283%2Ce%29.

3)  ln(x) = -1  ---->  x = e^(-1) = 1/e.

Solved (correctly).