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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.

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Are there any constraints on $x$ and $y$? Anyway, as a first step, one could simplify a little bit and obtain $\lceil \log(\lceil y/x\rceil) / \log (x-1) \rceil = 1$, or equivalently, $0 < \log(\lceil y/x\rceil) / \log(x-1) \le 1$. –  Johannes Kloos Apr 24 '12 at 7:07

3 Answers 3

up vote 2 down vote accepted

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 $$0 < \frac{\log \left\lceil \frac{y}{x} \right\rceil}{\log x-1} \le 1.$$

For now, assume $x > 2, y > 1$. Then $\log x - 1 \ge \log 1 = 0$, so $$0 < \log \left\lceil \frac{y}{x} \right\rceil \le \log (x-1).$$ Taking exponentials on both sides, we get $1 < \left\lceil \frac{y}{x} \right\rceil \le x-1$.

Using the definition of $\lceil \cdot \rceil$ again, we find that since $1 < \lceil y/x \rceil$, $1 < y/x$, 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 $x < y \le x^2 - x$. 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 $y \le 1$, $0 < \log\lceil y/x\rceil$ fails to hold, so the above lists all solutions for $x > 2$.

In the case that $1 < x \le 2$, 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 $0 < x-1 \le \lceil y/x\rceil < 1$. 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 $x > 2, x < y \le x^2-x$, but not every tuple of this form is a solution.

Edit: For a more exact solution, see Henry's post.

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$x < y \le x^2-x$ is satisfied by $x=e+1\approx 3.718, y=9$ but the original expression is not. –  Henry Apr 24 '12 at 8:07

$$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$

$$\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 1$$

$$0 \lt \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \le 1$$

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 $x \le 1$.

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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.

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