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.