How can we find or why can't we find the inverse function of $f(x)=\ln(x)+x$ for $x>0$?

Question

I want to find the inverse function of $f(x)=\ln(x)+x$, for $x>0$, but I heard it can't be done. Is there a way to find the inverse or to show it can't be found?

Answer 1

First of all, one shall start by finding the domain of the function and the domain of the inverse.

The function $f(x)$ is defined for such $x$, as : $$D_f = \{x\in \mathbb R : x>0\} = \mathbb R^+$$

Taking the first and second derivatives, we have :

$$f'(x) = \frac{1}{x} +1 >0, \; \; x>0$$

$$f''(x) = -\frac{1}{x^2} <0, \; \; x>0$$

Thus, the function $f$ is strictly increasing in its domain $D_f$ which also means that is $"1-1"$, which means that there exists an inverse function $f^{-1}(x)$ for $x>0$. Also take note that :

$$\lim_{x \to 0^+}f(x) = -\infty \; \; \text{and} \; \; \lim_{x \to \infty}f(x) = +\infty$$

Thus the range is $\mathbb R$ and the function $f$ is defined as $f: \mathbb R^+ \to \mathbb R$.

The inverse operation shall be carried out as finding an expression for $x$ involving the function family of values $y=f(x)$. Thus :

$$y=f(x) \Rightarrow y = \ln x + x $$

Now, take note that there is no possible way of solving this equation with respect to $x$. One cannot express $x$ simply and strictly in terms of $y$ and constants.

There are numerical methods and special function that handle such cases and for this particular example, the solution to the equation yielded is given as an expression of the product log function $\rm W$ :

$$y=\ln x + x \Leftrightarrow e^y = e^{\ln x + x} \Leftrightarrow e^y=xe^x \Leftrightarrow f^{-1}(x) = \rm W(e^x)$$

where $\rm W$ represents the Lambert $\rm W$ function. For more information about this special function, take a look here.

Finally, a graph showing both $f(x)$ and its inverse :