Question 487617
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            @ThePenguin solved the problem incorrectly, by incorrectly presenting the second solution and by loosing the third solution.


            I came to bring a correct version.



<pre>
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(3,e)}}}.

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

Solved (correctly).