How do you move floors and ceilings in discrete math?

[iScience]

The title more accurately should have been "How do you cancel floors and ceilings in discrete functions"

For instance,

$\frac{log{\frac{3x}{-6(z)}}}{8t} < 1$

If I wanted to get rid of the log, I'd just raise the expression by base 10.

$\frac{(\frac{3x}{-6(z)})}{10^{8t}} < 10^1$

But what happens if there's a roof for discrete functions?

$\frac{\lceil{log \frac{10x}{4y}}\rceil}{8z} < 1$

How do I handle this?

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EDIT NOTICE: the expressions above have been fixed into its their proper inequalities

 

[haruspex]

For instance,

$\frac{log{\frac{3x}{-6(z)}}}{8t}$

If I wanted to get rid of the log, I'd just raise the expression by base 10.

$\frac{(\frac{3x}{-6(z)})}{10^{8t}}$

i rather hope you would not do that, since it is wrong.

 

[iScience]

oops, sorry, here's the other side

$\frac{log(stuff)}{8t} < 1$

$\frac{stuff}{10^{8t}} < 10$

 

[haruspex]

oops, sorry, here's the other side

$\frac{log(stuff)}{8t} < 1$

$\frac{stuff}{10^{8t}} < 10$

Still wrong.

$\frac{\log(stuff)}{8t} < 1$
$\log(stuff)<8t$
$stuff<10^{8t}$
For your ceil question, it might help if you state the entire problem.

 

[mfb]

But what happens if there's a roof for discrete functions?

$\frac{\lceil{log \frac{10x}{4y}}\rceil}{8z} < 1$

Assuming z>0, z>0 leads to a very similar case:
$\lceil{stuff\rceil} < 8z$
What is the largest value of stuff that satisfies the inequality? Once you found that, you can continue with the usual approaches.