How do I show that $x=1$ is the only solution to $\textrm{e}^{x-1}=x$? How can I solve this equation: $e^{x-1}=x$ ?
I know that $x=1$ is a solution but how can I prove that it's the only one?
I can sketch each graph but I don't think it's adequate for a proof.
Thanks.
 A: Think about what your sketch looks like.  As $x$ increases, the exponential curve is increasing faster than the line, and so never comes back.  As $x$ decreases, below $1$, the curve flattens out, while the line just keeps decreasing, moving away from the curve.
This reasoning suggests a calculus based approach (we're using terms like "faster" and "flattening out" and "increasing").  Since we are looking for solutions of the equation $f(x)=0$, where $f(x)=\textrm{e}^{x-1}-x$, let's consider the derivative $f'(x)$:
$$f'(x)=\textrm{e}^{x-1}-1.$$  This function is clearly positive for $x>1$, and negative for $x<1$.  Importantly, $f$ has no turning points other than $x=1$!  
So, how do we use this observation to build a formal proof that $x=1$ is the only solution to $f(x)=0$?  Well, suppose that $f(x_{1})=0$ for some $x_1\neq 1$.  Then the average slope of $f$ between $0$ and $x_1$ would be $0$, which means that by the Mean Value Theorem, there would be some point $c$ between $1$ and $x_{1}$ for which $f'(c)=0$.  Since we know that there is no such $c$, we can conclude that there is no such $x_1$. 
