Solving $\log(x) = x-1$? One can use Taylor series of the log or exp function to get the result that $x = 1$. I was wondering if there is any other simple solutions.
Thanks a lot!
 A: The function $f(x) = x - \log x$ has derivative $f'(x) = 1 - \frac1x$ which is negative for $x<1$ and positive for $x>1$. Therefore, $f$ is strictly decreasing for $x<1$ and strictly increasing for $x>1$. Since $f(1)=1$ we have $f(x)>1$ for $x<1$ and $x>1$. Therefore, $x=1$ is the only solution to $f(x)=1$.
A: There is also an infinite set of complex solutions (if you already had complex numbers) if you consider different branches of the complex logarithm (Where $k$ denotes the branch):
$$\log(k, x) = x-1$$
$$x=e^{x-1}$$
$$xe^{1-x}=1$$
$$-xe^{-x}=-\frac{1}{e}$$
$$x = -W_k(-\frac{1}{e}) \land k \in \mathbb{Z}$$
where $W_k$ is the $k$-th branch of the $W$-Function.

Note that if you are taking the "main" branch of the logarithm in your equation, there is only one solution: $x = 1$
A: Suppose that $x>1$. Then $$\log x=\int_1^x t^{-1}dt <\int_1^x dt = x-1$$
Now suppose $0<y<1$. Then $$\log y=\int_1^y t^{-1}dt >\int_1^y t^{-2}dt=1-1/y$$
Now set $y=x^{-1}, x>1$ to get $\log x < 1-x$ on $0<x<1$. Equality is clear at $x=1$. In particular we have proven that for $x>0$ $$1-x^{-1}\leqslant \log x\leqslant x-1$$
with equality iff $x=1$. This proves in particular that $\log '1=1$, i.e. that $$\lim_{h\to 0 }\frac{\log(1+h)}h=1$$  
