Find number of zeros $ x = \ln |x-a| $ Find number of zeros $$ x = \ln |x-a| $$
depending on the value of $a$.
My try
Usually I solve task like that in use of derivatives: 
Let $$f(x) = x - \ln |x-a| $$
Ok now it is time for derivatives and check what is happening
Assume that $f'(x)$ exists... 
$$f'(x) = 1 - \frac{(|x-a|)'}{|x-a|} $$
Okay, I see that for $$|x-a| = 0 \rightarrow x=a$$
$f'(x)$ doesn't exists but it is not weird because $f(x)$ doesn't exists too. 
Now I can check different cases for $x<a$, $x>a+1$ and $x \le a+1$ - then I can see behavior of $f$ depending on the value of $a$. But there is some cases to check. I wonder if there is a simpler solution, maybe "tricky" one?
 A: The function $\ln|x-a|$ has two branches (the left branch which is strictly decreasing and the right branch which is strictly increasing). Since these branches are disjoint at $x=a$, we obtain$$\text{number of roots of }(x=\ln|x-a|){=\text{number of intersections of }y=x \text{ with the left branch} \\+\text{number of intersections of }y=x \text{ with the right branch}}$$but from the other side$$\text{number of intersections of }y=x \text{ with the left branch}=1$$which is quite obvious. Now define $$r(a)=\text{number of intersections of }y=x \text{ with the right branch}$$Different scenarios are considerable.
Case 1 : $a>-1$
In this case according to the famous inequality $u\ge\ln1+u$ we can write$$x{> x-a-1\\\ge \ln 1+(x-a-1)\\=\ln x-a\\=\ln|x-a|}$$therefore the right branch has no intersection with $y=x$ and $r(a)=0$.
Case 2 : $a=-1$
Once more, according to the inequality $u\ge \ln1+u$, the only intersection happens at $x=0$ therefore $r(a)=1$.
Case 3 : $a<-1$
In this scenario, since $\ln x-a$ is concave and $\ln (x-a)\Big |_{x=0}=\ln (-a)>1=\ln 0=x\Big|_{x=0}$ the there are exactly two intersections with $y=x$ hence $r(a)=2$.

Conclusion
The final formula for number of roots of $x-\ln|x-a|$
  is:$$r(a)=\begin{cases}1&,\quad a>-1\\2&,\quad a=-1\\3&,\quad
 a<-1\end{cases}$$

A: Graph $\ln (s)$ for $s > 0$.  An examination of the derivative w.r.t. $s$ shows that the function is strictly increasing, from $-\infty$ to $\infty$.  Therefore, there can be only one zero, which is known to be $s = 1$.
A: The function will either have one root, two or three roots. You must determine the local minimum $x^*$. If $f(x^*)<0$ there will be three roots, if $f(x^*)=0$ there will be two roots and if $f(x^*)>0$ there will be a single root.
Naturally, $x^*$ will depend on $a$. The local minimum will occur in $]a, +\infty[$, so
$$
f'(x^*)=0 \Leftrightarrow 1- \frac{1}{x^*-a} = 0 \Leftrightarrow \frac{x^*-a-1}{x^*-a} = 0 \Leftrightarrow x^* =1+a.
$$.
So, knowing the sign of $f(1+a)$ you can decide how many zeros there are. Since $f(a+1)=a+1$, there will be a single root if $a>-1$, two roots if $a=-1$ and three roots if $a< -1$.
A: This is a nice problem.  There is always one zero at positive $x$. But the answer is that there are $3$ zeros when $a<-1$, two of which come at opposite sides of the vertical line $x = a$, two zeros when $a = -1$ coming at $x\approx -1.278$ and $x=0$, and just the one positive one solution when $a>-1$.
A: The function $f(x)=x-\ln\lvert x-a\rvert$ is defined over $\mathbb{R}\setminus\{a\}$. Consider that
$$
\lim_{x\to-\infty}f(x)=-\infty,\quad
\lim_{x\to a^-}f(x)=\infty,\quad
\lim_{x\to a^+}f(x)=\infty,\quad
\lim_{x\to-\infty}f(x)=\infty
$$
and
$$
f'(x)=1-\frac{1}{x-a}=\frac{x-a-1}{x-a}
$$
For $x<a$, we also have $x<a+1$, so $f'(x)>0$. Hence, because of the limits above, there is a zero of $f$ in the interval $(-\infty,a)$.
For $x>a$, the function has a minimum at $a+1$. Since
$$
f(a+1)=a+1
$$
you should be able to finish.

 this minimum is negative for $a<-1$: two zeros; zero for $a=-1$: one zero; positive for $a>-1$: no zero.

You can also simplify the task by considering $g(x)=x+a-\ln\lvert x\rvert$, that has the same number of zeros. The derivative is $g'(x)=1-1/x=(x-1)/x$, so there is a minimum at $1$.
