Rearrange for T and J;

n=(2J+1)exp(-(BJ(J+1))/(KT))
where B and K are constants

Here's how you would do it for T.

ln n = ln (2J+1) -BJ(J+1)/kT

ln [n/(2J+1)] = -BJ(J+1)/kT

kT = -BJ(J+1)/ln[n/(2J+1)]
Now divide both sides by k

Work similarly to solve for J

I understand how you do it for T, but since there is more than one J, I don't understand how you make J the subject.

To rearrange the equation for T and J, we'll follow these steps:

Step 1: Remove the parentheses
n = (2J + 1) * exp(-(BJ(J + 1))/(KT))

Step 2: Distribute the exponent using the rule a * (b + c) = a * b + a * c
n = (2J + 1) * exp(-BJ^2 - BJ)/(KT)

Step 3: Move the exp(-(BJ^2))/(KT) term to the other side of the equation by dividing both sides by exp(-(BJ^2))/(KT)
n / exp(-(BJ^2))/(KT) = 2J + 1

Step 4: Simplify the exponential terms by multiplying the denominator with the exponent
n * exp(BJ^2)/(KT) = 2J + 1

Step 5: Rearrange the equation to isolate J on one side
n * exp(BJ^2)/(KT) - 1 = 2J

Step 6: Divide both sides by 2 to solve for J
J = (n * exp(BJ^2)/(KT) - 1) / 2

Step 7: Now, to isolate T, we can rearrange the equation as follows:

n = (2J + 1) * exp(-(BJ(J + 1))/(KT))

Step 8: Divide both sides by (2J + 1) to get rid of it on the right side
n / (2J + 1) = exp(-(BJ(J + 1))/(KT))

Step 9: Take the natural logarithm (ln) of both sides to cancel out the exponential function
ln(n / (2J + 1)) = ln(exp(-(BJ(J + 1))/(KT)))

Step 10: Simplify the right side by removing the natural logarithm and the exponential function
ln(n / (2J + 1)) = -(BJ(J + 1))/(KT)

Step 11: Multiply both sides by -(KT)/(BJ(J + 1))
-(KT)/(BJ(J + 1)) * ln(n / (2J + 1)) = -1

Step 12: Divide both sides by -1 to solve for T
T = (KT)/(BJ(J + 1)) * ln(n / (2J + 1))

Thus, the rearranged equations for T and J are:

J = (n * exp(BJ^2)/(KT) - 1) / 2

T = (KT)/(BJ(J + 1)) * ln(n / (2J + 1))