How to show $f(x)=(1+\ln(x))^2-2x$ has an unique zero analytically

Question

For the study of this function, I have to calculate the zeros of $f:\Bbb R_{>0}\to \Bbb R$ with $$f(x)=(1+\ln(x))^2-2x.$$ I was able to prove that it has a unique zero (and in fact it is in the interval $(0;e^{-1})$), that was found numerically. But I can not solve it analytically. Any ideas?

Answer 1

It can be shown using fairly elementary arguments: Consider the first derivative

$$ f'(x) = \frac{2 (1 + \log(x))}{x} - 2$$

and the second derivative

$$f''(x) = -\frac{2 \log(x)}{x^2}.$$

The second derivative has a zero at $x=1$. To the left side, on the interval $(0,1)$ the second derivative is positive as $\log(x) <0$. On the right side in the interval $(1,\infty)$ it is negative, as $\log(x)>0$.

This means the first derivative is monontonically increasing in $(0,1)$ and monotonically decreasing in $(1,\infty)$, and it is zero at $x=0$. So we can conclude

$$f'(x) < 0$$

for all $x\neq 0$, and $f'(x) = 0$ for $x=0$. This means $f$ is strictly monotonically decreasing almost everywhere ($x \neq 0$). This implies that there can be at most one zero.