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))