n=(2J+1)exp(-(BJ(J+1))/KT)
Rearrange for T and J; B and K are constants.
$\displaystyle n = (2j+1)e^{-\frac{bj(j+1)}{kt}}$
$\displaystyle \ln{n} = \ln(2j+1) + \left(-\frac{bj(j+1)}{kt}\right)$
$\displaystyle \ln\left(\frac{n}{2j+1}\right) = -\frac{bj(j+1)}{kt}
$
$\displaystyle t = -\frac{bj(j+1)}{k\ln\left(\frac{n}{2j+1}\right)}$
I may be mistaken, but I don't think there is an elementary solution for "j".
Maybe someone more knowledgeable can show us how.