Sanity Check: Solving for t in x=12(1-e^-t/RC) with Inversion Method

  • Thread starter parabol
  • Start date
That's not right. It should be 1/(1 - e^(-t/RC)).In summary, the solution to the given problem is t = -RC ln(1-x/12) and it is valid to invert both sides of the equation e^(-t/RC) = (12-x)/12 to get e^(t/RC) = 12/(12-x).
  • #1
parabol
17
0

Homework Statement



Transpose for t,

x=12(1-e^(-t/RC))

I can't get it out of my head but, the soltuion I have come up with doesn't seem right.

Solution is,

t=-RC ln(1-x/12)

Thanks in advance,

Realted, but not as important as the sanity check. For my own mind, could someone explain to me if this is possible and should/could be carried out. If during my transposing I came up with:

e^(-t/RC)= (12-x)/12

Is it possible to invert both sides so that the follwoing is shown

e^(t/RC)=12/(12-x)

Again, thanks in advance.
 
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  • #2
For your first problem, that's what I get.
For the second, yes, it's valid to invert both sides. The idea is that if a/b = c/d, then b/a = d/c, as long as you don't introduce any division by zero in doing this.
 
  • #3
Mark44 said:
For your first problem, that's what I get.
For the second, yes, it's valid to invert both sides. The idea is that if a/b = c/d, then b/a = d/c, as long as you don't introduce any division by zero in doing this.



Thanks Mark. Its just that I tried the invert first off and got completely different colutions once I had substitued values in (why I got confused) .
 
  • #4
I'm guessing that you made a mistake when you inverted first. If you have a sum or difference -- a + b -- the reciprocal isn't 1/a + 1/b. It's 1/(a + b). For your problem I'm guessing that you took the reciprocal of 1 - e^(-t/RC) to be 1 - e^(t/RC).
 

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