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Rearranging x=x0e^-lambda t in the form y=mx+c

  1. Mar 13, 2016 #1
    Hi, How could I rearrange x=x0e^-lambda t into the form y=mx+c, where y is equal to ln x and x is equal to t?

    Thank you in advance

    I tried to solve the problem myself, by taking the natural log of both sides, this left me with:

    ln x = -lambda t * lnx0

    However, I am not sure if this answer is correct or not

    If this is correct, would -lambda represent the gradient? And, would ln x0 represent the y intercept
     
    Last edited: Mar 13, 2016
  2. jcsd
  3. Mar 13, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    What is h0?
    The logarithm of a product is the sum of logarithms, so you should get ##-\lambda t + \ln(x_0)##.
    Sure.
     
  4. Mar 13, 2016 #3
    Thank you
     
  5. Mar 13, 2016 #4
    My mistake h0 was supposed to represent x0
     
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