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Exponentials Problem

  1. Jun 19, 2006 #1
    I have ln((V+v)/(V-v)) = 2ctV, and I've checked this is right. But how do I go from here to v = V((e^Vct - e^-Vct)/(e^Vct+e^-Vct))? The solutions just give them as successive lines.

    (I appreciate that's not the easiest thing to read, it's taken from the 2004 MEI specimen paper for A-level Mechanics 4, if that helps.)
  2. jcsd
  3. Jun 19, 2006 #2


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    HINT: Laws of logs;

    [tex]\ln\frac{a}{b} = \ln|a| - \ln|b|[/tex]
  4. Jun 19, 2006 #3
    OK, so
    ln((V+v)/(V-v)) = 2ctV
    => ln(V+v)-ln(V-v) = 2ctV
    ln(V)ln(v) - ln(V)/ln(v)
    2ln(v) = 2ctV
    v = e^Vct

    Is this right? Where can I go from here?
  5. Jun 20, 2006 #4
    there is something wrong in what u have written........ first of all
    log(a*b) = log(a) + log(b)...what u have used is log(a+b) = log(a) * log(b)..even after that u have written something wrong...check it once more....
    anyway u dont need to use that.....use the basic definition of logs.........
    if ln(a) = b
    => e^b = a
  6. Jun 21, 2006 #5


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    In other words,
    [tex]\frac{V+v}{V-v}= e^{2ctV}[/tex]
    Solve that for v.
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