Rearrangement?

Question

Been staring at this for ages and can't see how to do it any help appreciated thanks.

x = (m.s.e^(c.t))/(s - m+ m.e^(ct))

Simplified to
x = s.m/(m+(s - m).e^(ct))

Where all are constants except x and t. Thanks again sorry it messy (if it wasn't I would probably be able to do it, then again....)

Answer 1

Well, I don't see how that works exactly the way you posted it, but I can see how it works if there are either

1) two minuses in the original problem, which should have read:

x = (m × s × e^(-c × t)) / (s - m + m × e^(-c × t))

OR

2) one minus in the final answer, which should have read:

x = (s × m) / [m + (s - m) × e^(-c × t)]

I'll assume you posted the problem correctly but messed up the answer. The process is the same either way.

x = (m × s × e^(c × t)) / (s - m + m × e^(c × t))

Multiply top and bottom by e^(-c × t):

x = (m × s × e^(c × t)) / (s - m + m × e^(c × t)) × e^(-c × t)/e^(-c × t)

Distribute it through:

x = (m × s × e^(c × t) × e^(-c × t)) / [(s - m + m × e^(c × t)) × e^(-c × t)]

And then e^(c × t) × e^(-c × t) = e^(c × t - c × t) = e^0:

x = (m × s × e^0) / [s × e^(-c × t) - m × e^(-c × t) + m × e^0]

And of course e^0 is just 1:

x = (s × m) / [m + (s - m) × e^(-c × t)]

If the minuses should be in the original equation instead of the final solution, then you'd multiply by e^(c × t)/e^(c × t), instead.

Would be interested to know which it was.... If it's neither, then the person who wrote it messed up.

- - - - - - - - - - - -

Cliff's Notes Version: n = ct

           mseⁿ           eˉⁿ              ms
x = - - - - - - - - -  ×  - - - = - - - - - - - - - - -
     s - m + meⁿ      eˉⁿ     seˉⁿ - meˉⁿ + m

             ms
x = - - - - - - - - - - -
      m + eˉⁿ(s - m)

Answer 2

This is a horse with a broken leg, and needs to be shot.

This is totally messed up, and the original expression really can't be simplified any further anyway, if you've posted it correctly.