Isolate t from x(t)=vτ(1-e-t/τ) Homework

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To isolate t from the equation x(t) = vτ(1-e^(-t/τ)), first rearrange the equation to isolate the exponential term. Subtract 1 from both sides and then multiply by -1 to obtain e^(-t/τ) = 1 - (x(t)/vτ). Next, take the natural logarithm of both sides, leading to -t/τ = ln(1 - (x(t)/vτ)). Finally, solve for t by multiplying both sides by -τ. This process effectively isolates t in the equation.
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Homework Statement


I just need to isolate t from the equation though I'm stumped on how to do it.

Homework Equations



x(t) = vτ(1-e-t/τ)

*τ and v are constants

The Attempt at a Solution


I know you somehow need to take the ln of both sides so

ln x(t)/vτ = ln (1-e-t/τ)

That's pretty much where I got to so far. Any suggestions on how to proceed to the next steps will be appreciated. Thanks!
 
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welcome to pf!

hi zeromaxxx! welcome to pf! :smile:
zeromaxxx said:
x(t) = vτ(1-e-t/τ)

*τ and v are constants

I know you somehow need to take the ln of both sides …

yes, "somehow" is the important word …

the trick is to get the e on its own on one side, and everything else on the other …

then you have ln(e) on one side, which is simple

(and not ln(1 - e)), which is useless! :wink:)
 
zeromaxxx said:

Homework Statement


I just need to isolate t from the equation though I'm stumped on how to do it.

Homework Equations



x(t) = vτ(1-e-t/τ)

*τ and v are constants

The Attempt at a Solution


I know you somehow need to take the ln of both sides so

ln x(t)/vτ = ln (1-e-t/τ)
Too soon! First isolate the exponential by (1) subtract 1 from both sides, (2) multiply both sides by -1:
e^{-t/\tau}= 1- \frac{x(t)}{v_T}

NOW take the natural logarithm of both sides.


That's pretty much where I got to so far. Any suggestions on how to proceed to the next steps will be appreciated. Thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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