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)   (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(52781)   (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) = . 3) ln(x) = -1 ----> x = e^(-1) = 1/e.

Solved (correctly).