- #1

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V1 = Vf(1-e^(-t1/tau)) and V2 = Vf=(1-e^(-t2/tau))

I need to solve for tau in terms of t1, t2, Vf, V1, and V2 but i cant seem to find a proper way to combine these equations.

can you help?

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- Thread starter Mugged
- Start date

- #1

- 104

- 0

V1 = Vf(1-e^(-t1/tau)) and V2 = Vf=(1-e^(-t2/tau))

I need to solve for tau in terms of t1, t2, Vf, V1, and V2 but i cant seem to find a proper way to combine these equations.

can you help?

- #2

cepheid

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Example of solving one of the equations:

[tex] V_1 = V_f (1 - e^{-t_1 / \tau} ) [/tex]

Divide both sides by V

[tex] \frac{V_1}{V_f} = 1 - e^{-t_1 / \tau} [/tex]

Subtract V

[tex] e^{-t_1 / \tau} = 1 - \frac{V_1}{V_f} [/tex]

Take the natural logarithm of both sides of the equation:

[tex] -\frac{t_1}{\tau} = \ln \left[1 - \frac{V_1}{V_f} \right] [/tex]

Solve for tau by multiplying both sides by tau and then dividing both sides by the ln term (in other words, cross-multiply):

[tex] \tau = -\frac{t_1}{\ln \left[1 - \frac{V_1}{V_f} \right]} [/tex]

Now, when you solve the second equation for tau, you'll get a similar answer in terms of V

- #3

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I shouldve made that clear; not hard to solve for tau using just 1 equation actually...

- #4

cepheid

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I shouldve made that clear; not hard to solve for tau using just 1 equation actually...

Yeah, I know that. You should have read my post more closely. Do you agree with me, that using the exact same algebraic steps described in my first post, but applying them to the second equation (for V

[tex] \tau = -\frac{t_2}{\ln \left[1 - \frac{V_2}{V_f} \right]} [/tex]

If so, then it must be true that:

[tex] \tau = -\frac{t_2}{\ln \left[1 - \frac{V_2}{V_f} \right]} = -\frac{t_1}{\ln \left[1 - \frac{V_1}{V_f} \right]} [/tex]

EDIT: So basically V2, t2, V1, t1, can't all be independent variables. The equation above illustrates that they must be related to each other.

Last edited:

- #5

- 104

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actually i already solved it myself just now. if you solve for the e pieces and divide one over the other, you can solve for tau in terms of all variables.

thanks though ceph

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