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ODEs and ln function

  1. Mar 15, 2009 #1
    I don't understand what is happening in the following problem.
    What happens to ln when it moves to the RHS?
    Why are there two exponential functions on the RHS and why is e^c made to equal A?
    http://users.on.net/~rohanlal/mathproblemln.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 16, 2009 #2

    djeitnstine

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    c is a constant, so what does that make [tex]e^c[/tex]? Isn't that just some other number? Couldn't you just call that other number A? Also I do not see 2 exponential functions, I only see one...
     
  4. Mar 16, 2009 #3

    HallsofIvy

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    ln doesn't "move" to the right hand side! You should have learned not to use "baby" language like that long ago. If you have an operation on the left side of an equation that you want to eliminate, you do the opposite or inverse operation:
    1) if 2 is added to, the inverse of "add 2" is "subtract 2": x+2= 1 becomes x+ 2- 2= 1- 2 or x= -1.
    2) if 3 is multiplied by x, the inverse of "multiply by 3 is "divide by 3": 3x= 1 becomes (3x)/3= 1/3 or x= 1/3.

    3) If the problem is ln(x)= 1, you do the inverse of ln. What is the inverse of y= ln(x)?
     
  5. Mar 17, 2009 #4
    e to the power of c i'd assume.But still that leaves the question of why there is a 2nd e
    and why it is raised to the power of all of the terms that were initially on the RHS.
     
  6. Mar 17, 2009 #5

    HallsofIvy

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    Yes, the inverse of y= ln(x) is y= ex (not "c").

    So if ln(x)= a, what is x?
     
  7. Mar 17, 2009 #6
    well a=e^x
    so therefore aloge=x
    but that still doesn't answer my question
     
  8. Mar 17, 2009 #7
    Maybe this will clarify things:

    [tex]e^{\frac{t^2}{2}+2t+c}=e^{c}e^{\frac{t^2}{2}+2t}=Ae^{\frac{t^2}{2}+2t}[/tex]

    where I used that [tex]e^{a+b}=e^ae^b[/tex] and [tex]A=e^c[/tex] by definition.
     
  9. Mar 24, 2009 #8
    why doesnt the resulting equation
    look like Ae*(1/2)*(t^2)+2t
    rather than Ae^(1/2)*(t^2)+2t
     
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