polynomial equation degree 3 with logarithm I have a polynomial equation of third degree 
$$x^{3}+2x^{2}+\log(x)-4=0$$
Since there is a term with $\log$, I do not know how to manage it to find a solution.
Is there any way to solve this equation and find the corresponding three roots for this equation?
 A: Let $f(x)=x^{3}+2\,x^{2}+\log(x)-4$. It is clear that $f$ is continuous on $(0,+\infty)$. We have $\lim_{x\to0^+}f(x)=-\infty$ and $\lim_{x\to+\infty}f(x)=-+\infty$. By the intermediate value theorem there is at least one $x>0$ such that $f(x)=0$. Moreover $f'(x)=3\,x^2+4\,x+1/x>0$, so that $f$ is strictly increasing. This implies that there is a unique solution to the equation $f(x)=0$. To locate it, give values to $x$: $f(1)=-2<0$, $f(2)=12+\log2>0$. The root is between $1$ and $2$. You can use now your favorite numerical method to approximate the root, like the bisection method, the secant method or Newton's method. Or you can use software, like SAGE or Wolfram alpha.
A: The extrema of the functions are found by cancelling the derivative
$$3x^2+4x+\frac1x=0.$$
This function remains positive for $x>0$, so that the initial function is growing can only have one root.
And there is indeed a root between $1$ and $2$ as $f(1)f(2)<0$, which you need to refine numerically.
A: For example, here is a python script that uses the secant method:
from math import log
def f(x):
    return x*x*(x+2)+log(x) -4

epsilon = 1.0e-8
x=[1,1.5]
while abs(x[-1]-x[-2])> epsilon:
    x1,x2 = x[-2], x[-1]
    x.append(x2-f(x2)*(x2-x1)/(f(x2)-f(x1)))

print(x[-1])

The output is 
1.117033697609942

