Solve for $x$ in $x\ln(2-x)\ge 2x$ $x\ln(2-x)\ge 2x$
I did:
$$x\ln(2-x)\ge 2x \Longleftrightarrow \ln((2-x)^x) \ge 2x \Longleftrightarrow e^{2x}\ge (2-x)^x \Longleftrightarrow \text{???}$$
What do I do next? Am I doing it correctly?
 A: for $x=0$ we have $$0\geq 0$$ which is true. For $x>0$ we have to solve $\ln(2-x)\geq 2$ and for $x<0$ we have to solve $\ln(2-x)\le 2$
A: Simply dividing both sides by $x$ is not valid unless $x>0$, so you need to check separately whether $x=0$ is a solution. And it is. But if $x<0$, then you need to reverse the direction of the inequality when you divide both sides by $x$.
Note that $\ln((2-x)^x) = x\ln(2-x).$
So where you have $\ln((2-x)^x) \ge 2x,$ you can write $x\ln(2-x)\ge 2x$ and then divide both sides by $x$ again, getting $\ln(2-x)\ge2$ if $x>0$.  From that you get $2-x \ge e^2,$ and finally $x\le e^2-2.$
Also, notice that with your more complicated version that says $(2-x)^x \ge e^{2x},$ you can raise both sides to the power $1/x$, getting $2-x \ge e^2,$ and then solve that for $x$.
A: The inequality only makes sense for $x<2$. You have to find where
$$
x(\ln(2-x)-2)\ge0
$$
which happens when
$$
\begin{cases}
x<2\\[4px]
x\ge0 \\[4px]
\ln(2-x)-2\ge0
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x<2 \\[4px]
x\le0 \\[4px]
\ln(2-x)-2\le0
\end{cases}
$$
This becomes
$$
\begin{cases}
0\le x<2 \\[4px]
2-x\ge e^2
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x\le 0 \\[4px]
2-x<e^2
\end{cases}
$$
A: Well, consider the domains of both sides. The domain of $2x$ is $\mathbb{R}$, while the domain of $x \ln(2-x)$ is $x<2$. This is easy to see because the unextended logarithm is only defined for $[0,\infty) \setminus  0^+$. 
Then solution 1: $x = 0$
Then following where you were:
$x\ln(x) \ge 2x$
$\exp(\ln(2-x)) = 2-x \ge \exp(2)$
$x \le 2 - \exp(2) \approx -5.38...$
So the solution is $x=0$ and $x < 2-\exp(2)$
