Solving an equation with nested ceilings How would I go about solving something like the following for $x$ in terms of $y$?
$$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$
I've never had to solve an equation with a ceiling before, never mind nested ceilings. Even Wolfram Alpha Pro is timing out to the point of not being able to go anywhere.
 A: $$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$
$$\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 1$$
$$0 \lt \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right )  \le 1$$
assuming $x \gt 2$:
$$1 \lt  \left \lceil \frac{y}{x} \right \rceil \le x-1$$ 
$$1 \lt   \frac{y}{x} \le \lfloor x-1 \rfloor$$
$$x \lt  y \le x \lfloor x-1 \rfloor.$$
Note if $1 \lt x \lt 2$ then you need to solve $x-1 \le  \left \lceil \frac{y}{x} \right \rceil \lt 1$, which would require you to find an integer between $0$ and $1$. You may have problems with the logarithm base $x-1$ if $x=2$ or $x \le 1$. 
A: First of all, simplify to $$\left\lceil \frac{\log(\left\lceil \frac{y}{x} \right\rceil)}{\log x-1} \right\rceil = 1.$$ Without loss of generality, $\log$ denotes the natural logarithm. Using the definition of $\lceil \cdot \rceil$, one finds $$0 &lt \frac{\log \left\lceil \frac{y}{x} \right\rceil}{\log x-1} \le 1.$$
For now, assume $x > 2, y > 1$. Then $\log x - 1 \ge \log 1 = 0$, so $$0 &lt \log \left\lceil \frac{y}{x} \right\rceil \le \log (x-1).$$ Taking exponentials on both sides, we get $1 &lt \left\lceil \frac{y}{x} \right\rceil \le x-1$.
Using the definition of $\lceil \cdot \rceil$ again, we find that since $1 &lt \lceil y/x \rceil$, $1 &lt y/x$, so $y > x$. Also, since $\lceil y/x \rceil \le x-1$, we have $y/x \le x-1$ as well, i.e., $y \le x^2-x$. In other words, solutions with $x > 2, y > 1$ will always be of the form $x &lt y \le x^2 - x$. Edit: Not all of these tuples fulfill the equation; you will need to check for every concrete pair.
Edit: If $x > 2$, it is also easy to see that if $y \le 1$, $0 &lt \log\lceil y/x\rceil$ fails to hold, so the above lists all solutions for $x > 2$.
In the case that $1 &lt x \le 2$, a similar calculation yields $0 > \log \lceil \frac{y}{x}\rceil \ge \log x-1$, so $1 > \left\lceil\frac{y}{x}\right\rceil \ge x-1$. By choise of $x$, we have $0 &lt x-1 \le \lceil y/x\rceil &lt 1$. Since $\lceil y/x \rceil$ is an integer, no solution exists in this case.
Finally, for $x \le 1$, $\log_{x-1}$ is not defined over $\mathbb R$ anymore, so these cases can be excluded.
All in all, the solutions to this equation are of the form $x > 2, x &lt y \le x^2-x$, but not every tuple of this form is a solution.
Edit: For a more exact solution, see Henry's post.
A: You realize that there is not necessarily a unique solution for this problem, right?
For instance, for any $\alpha \in [0,1]$:
$$y=x(x-1)^\alpha$$
Is a solution for some range of $x$. I'm sure there are many many more functions that satisfy your relation as well. Without minimal constraints on the function $y$ (continuous?, differentiable?), I don't think this is solvable.
