How to apply the floor function in algebra? I was reading a statistics paper and saw a formula with floor operators. I wondered how to solve for one of the variables in the formula, but realized that I did not know how to work with these things.
I have the following, where I am trying to solve for $a$.
$y = \lfloor log(a) - log(b) \rfloor + 1$
I am confused about the order of operations with the floor operators $\lfloor \rfloor$.
I see some options

*

*Ignore the floor and solve for $a$
$y = log(a) - log(b) + 1$
$y - 1 + log(b) = log(a)$
$a = exp[y - 1 + log(b) ]$


*Apply floor to each expression in the formula then use ceiling

$y = \lfloor log(a) - log(b) \rfloor + 1$
$y = \lfloor log(a) \rfloor - \lfloor log(b) \rfloor + 1$
$\lfloor log(a) \rfloor = y - 1  + \lfloor log(b) \rfloor$
$log(a)  = \lceil y  -  1   + \lfloor log(b) \rfloor\rceil$
$a  = exp[\lceil y  -  1   + \lfloor log(b) \rfloor\rceil]$
I could keep going with more I think, but I hope this gets the point across. I don't know how to do algebra with these ceiling and floor operators.

You have that $m = \lfloor{x}\rfloor$ if and only if $m \leq x < m + 1$. Thus,
$$
   y - 1 = \lfloor \log(a) - \log(b) \rfloor \implies y - 1 \leq \log(a) - \log(b) < y
$$
and so we get
$$
   y - 1 + \log(b) \leq \log(a) < y + \log(b)
$$
Exponentiating
$$
   be^{y - 1} \leq a < be^y
$$
and so on...
