Find out how many solutions has an equation $$ x=\ln(|x-a|) $$ depending on the value of $a$
My approach:
I see that as $\exp$ is an injective function the equation can be reduced to $$ e^x=|x-a| $$ and I also know that the line $y=1+x$ is tangent to function $e^x$ in $x=0$ so I would say that the initial equation has
- solution for $a\in(-1,\infty)$
- solutions for $x=-1$
- solutions for $x\in(-\infty,-1)$
but I don't know how to prove this facts formally using the real-analysis theorems.
Let $f(x) := |x-a|e^{-x}$. Then $x$ is a solution of your equation iff $f(x) = 1$. We have $f(x)\ge 0$ for all $x$, $f(a) = 0$, $f(-\infty) = +\infty$ and $f(+\infty) = 0$. On $(-\infty,a)$ we have $f(x) = (a-x)e^{-1}$ and thus $f'(x) = -(1+a-x)e^{-1} < 0$. Therefore, there is always exactly one solution in $(-\infty,a)$. Consider $f$ on $(a,\infty)$. There we have $f(x) = (x-a)e^{-x}$ with derivative $f'(x) = (1-(x-a))e^{-x}$. This one has a zero exactly at $x = 1+a$ and $f(1+a) = e^{-(1+a)}$ (here we have a maximum!). So, $f(1+a) < 1$ iff $a> -1$. This shows that you are right.