Prove that $ f(x) = e^x + \ln x $ attains every real number as its value exactly once

Prove that the function $$ f(x) = e^x + \ln x $$ attains every real number as its value exactly once.

First, I thought to prove that this function is a monotonic continuous function.

But then I wasn’t sure if that is how to prove the result, and if it is, I wasn’t sure of how exactly to prove it that way.

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Clearly, $f$ is continuous and differentiable over the set of positive numbers.

$$f'(x)=e^x+\frac{1}{x}$$
If $x>0, f'(x)>0$, hence it is monotonic continuous function.
Also, note that $$\lim_{x \rightarrow 0^+}f(x)=-\infty$$ and $$\lim_{x \rightarrow \infty}f(x)=\infty$$

Hence, it attains every real number exactly once.

You can prove it by noting three things:

  1. The function $f \colon (0,\infty) \rightarrow \mathbb{R}$ is continuous and strictly monotonic as
    $$ f'(x) = e^x + \frac{1}{x} > 0 \,\,\, \forall x \in (0,\infty). $$
  2. When $x \to 0^{+}$, we have
    $$ \lim_{x \to 0^{+}} f(x) = -\infty. $$
  3. When $x \to \infty$ we have
    $$ \lim_{x \to \infty} f(x) = +\infty. $$

By using $(2)$ and $(3)$, given $y \in \mathbb{R}$ we can find $x_0 < x_1$ such that $f(x_0) < y – \frac{1}{2}$ and $f(x_1) > y + \frac{1}{2}$. By continuity on $[x_0,x_1]$, $f$ must assume all values between $f(x_0)$ and $f(x_1)$ and in particular $y$. By strict monotonicity, it will assume $y$ only once.


Without proving $(2)$ and $(3)$, the function could miss some values if it has an asymptote at infinity or at $0$. For example, $f(x) = \arctan(x)$ is monotonic on $(0,\infty)$ but its image is $(0,\frac{\pi}{2})$.