How do I solve this equation with ln(x)?

Question

$$1 - \dfrac{1}{x} + \ln x = 0$$

I know that the solution is $1$, because I started from the assumption that $ \ln(1)=0$. However, we cannot take this assumption for granted.

So what are the steps to solve this logarithmic function?

Answer 1

Let $f(x) = 1-{1 \over x} + \log x$ on $(0,\infty)$.

Note that $f'(x) >0$, $\lim_{x \downarrow 0} f(x) = -\infty$, $\lim_{x \to + \infty} f(x) = \infty$, so we know there is a unique solution.

In this case, we can try some standard values such as $f(1) = 0$. Hence $1$ is the unique solution.

Answer 2

$$1+ \ln x = \frac{1}{x}$$

If $x>1$, $1+\ln x > 1$ but $\frac{1}{x} <1$.

If $x<1$, $1+\ln x<1$ but $\frac{1}{x}>1$.

Hence $x=1$ is the unique solution.

Answer 3

Your equation is equivalent to $e^{-e}=(xe)^{-xe}$ .

For $y=z^{1/z}$ there is only one solution for $0<z<1$ because $(z^{1/z})'=z^{(1/z)-2}(1-\ln z)>0$ for $0<z<1$ means $z^{1/z}$ is strictly increasing for $0<z<1$.

With $y:=e^{-e}$ and $z:=(xe)^{-xe}$ one gets only one solution for $x$ which is obviously $1$.