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E^-x, solve for x

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve for x

    0.5y = e^-x

    3. The attempt at a solution

    I believe the answer is x = ln(2/y)

    Would anyone be able to explain the intermediate steps please?

    If I was trying it I would have got -ln0.5y = x
     
  2. jcsd
  3. Dec 3, 2011 #2

    Curious3141

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    Homework Helper

    Your answer is right, and you can easily rearrange it to ln (2/y).

    What is the relationship between -ln (a) and ln (a)?
     
  4. Dec 3, 2011 #3
    If you take the log of either side you get

    [itex]ln(y/2)=-x[/itex] or rearranging terms and using [itex]ln(a/b)=ln(a)-ln(b)[/itex] gives [itex]x=-ln(y/2)=-(ln(y)-ln(2))=ln(2/y)[/itex]
     
  5. Dec 3, 2011 #4

    Mentallic

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    Homework Helper

    Or more simply,

    [tex]a\cdot \ln(b)=\ln(b^a)[/tex]

    so

    [tex]-\ln(y)=\ln(y^{-1})=\ln(1/y)[/tex]
     
  6. Dec 3, 2011 #5
    Thanks very much guys! :smile:
     
  7. Dec 3, 2011 #6

    Mark44

    Staff: Mentor

    Or, -ln(y) = 0 - ln(y) = ln(1) - ln(y) = ln(1/y).

    Here, I'm using the property that ln(a/b) = ln(a) - ln(b) (in reverse).
     
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