It has been 50 years since Algebra, could someone help me transpose an equation?

Question

The equation is T= (9*h*Log(Rb/Ra) / 8*pi^2*N^2*r^2* (p-po))+ 2*(ta+td)/3 I wish to transpose and solve for N in terms of all the other variables.

Ironduke, Yes the symbol in question is a divide. I expressed the equation in "basic" notation as I don't yet know how to use symbols while composing these messages. It looks as if the three of you agree, at a glance. I thank you all!

Answer 1

T= (9*h*Log(Rb/Ra) / 8*pi^2*N^2*r^2* (p-po)) + 2*(ta+td)/3
Subtract 2*(ta+td)/3:
T - 2*(ta+td)/3 = (9*h*Log(Rb/Ra) / 8*pi^2*N^2*r^2* (p-po))
Take reciprocals:
1 / ( T - 2*(ta+td)/3 )
= 8*pi^2*N^2*r^2* (p-po) / (9*h*Log(Rb/Ra)
Multiply by 9*h*Log(Rb/Ra):
(9*h*Log(Rb/Ra) / ( T - 2*(ta+td)/3 ) = 8*pi^2*N^2*r^2* (p-po)
Divide by 8*pi^2*r^2* (p-po):
N^2 = (9*h*Log(Rb/Ra) / ( T - 2*(ta+td)/3 )8*pi^2*r^2* (p-po)
Take square root:
N = +/- sqrt( (9*h*Log(Rb/Ra) / ( T - 2*(ta+td)/3 )8*pi^2*r^2* (p-po) ).

Answer 2

first you must determine the terms that contain N.
there just one term and contain one N.

then it is easy:

T - 2*(ta+td)/3 = [9*h*log(Rb/Ra)/8*pi*r^2*(p-po )] * N^2
N^2 = [ T - 2*(ta+td)/3 ] / [9*h*log(Rb/Ra)/8*pi*r^2*(p-po )]

then take the square root for the two sides, and the equation is solved for N.

Answer 3

T= (9*h*Log(Rb/Ra) / 8*pi^2*N^2*r^2* (p-po))+ 2*(ta+td)/3
T- 2(ta+td)/3= (9*h*Log(Rb/Ra) / 8*pi^2*N^2*r^2* (p-po))
Now is the right hand side of the above equation =
9hlog(Rb/Ra)
-------------------------- ?
8pi^2N^2r^2(p-po)
Your parentheses have me a bit confused. If the right hand side is what I have above, then:
8pi^2N^2r^2(p-po) = [9hlog(Rb/Ra)]/[T- 2(ta+tda)/3]
N^2 ={[9hlog(Rb/Ra)]/[T- 2(ta+tda)/3]}/(8pi^2r^2(p-po))
N is then just the sqrt of the above.