Number of solutions of $x+\log\vert x\vert =a$ for $a\in\mathbb{R}$

Question

Let $a\in\mathbb{R}$. I'm trying to study the number of solutions of the equation $$ x+\log\vert x\vert =a. $$ Plotting the function $x+\log\vert x\vert$ I can immediately see (graphically) that if $a>-1$ then we have a unique solution, whereas if $a<-1$ we have 3 solutions. The case $a=-1$ has exactly two solutions. However I've not been able to write this mathematically rigorous. I've been thinking if there is an “easy” way / intuitive way of proving these properties without relying in using wolfram to plot the function. First I thought splitting the cases $x\geq 0$ and $x<0$ (since we have an absolute value), but I am a little bit lost to be honest. It seems to me that the best option is trying to analyze the image of the function $f(x):(-\infty,0)\to \mathbb{R}$ where $f(x):=x+\log\vert x\vert$. Then, I would combine this with the fact that the image of the same function defined on $(0,+\infty)$ is the whole $\mathbb{R}$ and that $f(x)$ is strictly increasing there (so for any $a\in\mathbb{R}$ there is exactly one solution $x\in(0,+\infty)$ in this interval). Does this make sense right? However, I don't know how to exactly calculate what is the image of $f(x)$ on $(-\infty,0)$. Does anyone has any hint for this?

Answer 1

For each $x\ne0$ let$$f(x)=x+\log|x|=\begin{cases}x+\log x&\text{ if }x>0\\x-\log x&\text{ if }x<0.\end{cases}$$Then $f$ is differentiable and$$(\forall x\in\Bbb R\setminus\{0\}):f'(x)=1+\frac1x.$$So, $f$ is strictly increasing on $(-\infty,-1]$ and on $(0,\infty)$ and strictly decreasing on $[-1,0)$. Besides, $f(-1)=-1$ and, since$$\lim_{x\to0^+}f(x)=-\infty\quad\text{and}\quad\lim_{x\to\infty}f(x)=\infty,$$you have that, for each $y\in\Bbb R$, the equation $f(x)=y$ has one and only one solution in $(0,\infty)$. Since$$\lim_{x\to-\infty}f(x)=\lim_{x\to0^-}f(x)=-\infty$$ and since the maximum of $f|_{(-\infty,0)}$ is $-1$, attained at $-1$, then, for each $y\in\Bbb R$, the equation $f(x)=y$ has

• no roots in $(-\infty,0)$ if $y>-1$;
• exactly one root in $(-\infty,0)$ if $y=-1$;
• exactly two roots in $(-\infty,0)$ if $y<-1$.