How Can I Solve for Tau in These Two Equations?

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Homework Help Overview

The original poster presents two equations involving exponential decay, specifically V1 = Vf(1-e^(-t1/tau)) and V2 = Vf(1-e^(-t2/tau)), seeking to solve for tau in terms of five variables: t1, t2, Vf, V1, and V2.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss solving each equation for tau individually and equating the results to establish a relationship among the variables. There is a focus on ensuring tau is expressed in terms of all five variables rather than just three.

Discussion Status

Some participants have provided guidance on manipulating the equations to isolate tau, while others express a need for clarity on how to incorporate all five variables into the solution. The discussion reflects a mix of interpretations regarding the relationships between the variables.

Contextual Notes

There is an indication that the variables V1, V2, t1, and t2 cannot be treated as completely independent, suggesting a deeper relationship that needs to be explored in the context of the equations provided.

Mugged
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I have these two equations:

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 can't seem to find a proper way to combine these equations.

can you help?
 
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You can solve each equation individually for tau. Then, since the results must be equal, you can equate them. This will give you a relationship between V1, t1, and V2, t2.

Example of solving one of the equations:

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

Divide both sides by Vf:

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

Subtract V1 / Vf from both sides. Then add the exponential term to both sides (in other words, rearrange the terms):

[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 V2 and t2. Since it must be true that tau = tau, you can equate these results.
 
erm, thanks, but i need tau in terms of all 5 variables, not just 3.

I shouldve made that clear; not hard to solve for tau using just 1 equation actually...
 
Mugged said:
erm, thanks, but i need tau in terms of all 5 variables, not just 3.

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 V2), we will arrive at the following expression for tau?

[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:
yeah that's mathematically correct no doubt...but definitely not what i asked.

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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