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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 1.2012-04-24

3 Answers 3

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

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

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

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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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.2012-04-24