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.