Problem:
Find a curve whose slope at each point $(x,y)$ equals the reciprocal of the $x$-value if the curve contains the point $(e,-3)$.
If we call the equation of the curve $f(x)$, then $f'(x)=\frac{1}{x}$. $\int \frac{1}{x} \, dx = \ln|x|+C$, so $f(x)= \ln|x|+C$. $f(e)=-3$ so $f(x)= \ln|x|-4$. What I am confused by is that my book's answer is $f(x)= \ln x-4$ (there are no absolute value bars around $x$). If there are no bars, that means the bars are uneccessary so $x$ must be greater than zero. But what part of the problem indicates that $x$ must be greater than zero?
Since $\frac{1}{x}$ is discontinuous (update: Hans Lundmark's comment suggests using "undefined" instead) at $x = 0$, there are actually $2$ curves involved, one where $x \lt 0$ and another where $x \gt 0$. Since the point in question of $(e,-3)$ has $x = e \gt 0$, it's only this positive $x$-values curve which is being asked about. Thus, there's no need for absolute value signs in this case. Nonetheless, I believe the book should still have used absolute values, even if they're not strictly necessary.
Update: After reading the comments below and on further reflection, I now believe it's best to not use absolute values, as this helps to implicitly show the solution is only for $x \gt 0$. I suspect this is a major reason why the book did not use absolute values in their answer.