How do you solve $e^x=ex$? I mean the answer is really simple. It's one but how do you solve this equation?
 A: Since $e^x=e+e(x-1)+o(x-1)$ as $x \to 1$, the straight line $y=e+e(x-1)=ex$ is tangent to the graph $y=e^x$. By convexity, the exponential function is always above this tangent line, and hence $x=1$ is the only possible intersection.
A: First show that $x = 1$ is a solution, and then show that there is at most one solution. For the latter part of the proof, it helps to notice that the curves $y = e^x$ and $y = ex$ are tangent to each other at $x = 1$.
A: This is equivalent to $e^{x-1}=x$ after division by $e$. There is no algorithmic way of handling equations of this form. However, if you try a $x=1$, then you will see that $e^{1-1}-1=e^{0} - 1 = 0$.
If you would like to demonstrate that this solution is unique, then you may use some calculus. Notice that $f(x) = e^{x-1}-x$ has a strictly positive second derivative, $f''(x) = e^{x-1}$, and that $f'(x) = 0$ at $x=1$. Therefore $f$ has a unique minimum at $x=1$. Thus $f(x) = 0$ only happens when $x=1$ and that is the only solution to $e^{x-1}=x$.
A: You can see that $f(x)=e^x-ex$ has a minimum value at $x=1$ since $f'(x)=0$ has only one solution at $x=1$. 
