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Natural numbers [please help]

  1. Apr 21, 2005 #1
    Hello. I am working on this problem

    0=0.002*e^-(0.005/2R)

    I am supposed to find to find "R". The only way I know how to do this gives me 0... but I know that it's not the answer. Got any tips?
     
  2. jcsd
  3. Apr 21, 2005 #2

    James R

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    There is no solution.
     
  4. Apr 21, 2005 #3
    e^(x) > 0 for every x. This should clarify the point made by James R
     
  5. Apr 21, 2005 #4
    What if the number is small (but not zero)... ie 0.000001 = 0005e^(0.004/2R)
     
  6. Apr 22, 2005 #5

    HallsofIvy

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    Then it's easy (and NOT "Calculus and Analysis"!). Divide both sides by 0.0005 to get [tex]e^{\frac{0.004}{2R}}= \frac{0.000001}{0.0005}[/tex].

    Take the natural logarithm of both sides to get rid of the exponential:
    [tex]\frac{0.004}{2R}= ln(\frac{0.000001}{0.0005})[/tex]

    Multiply both sides by R:
    [tex]0.002= R ln(\frac{0.000001}{0.0005})[/tex]
    and, finally, divide both sides by the logarithm:

    [itex] R= \frac{0.002}{ln(\frac{0.000001}{0.0005}}[/itex]
     
    Last edited: Apr 22, 2005
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