Solving $\log(x) = x-1$?

Question

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!

Answer 1

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 $f(x)=1$.

Answer 2

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: $x = 1$

Answer 3

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 $$\lim_{h\to 0 }\frac{\log(1+h)}h=1$$