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Been trying to figure this one out for some time

  1. Oct 6, 2009 #1
    LogeX=2loge(100-y^2)+C where X(0)=100 find C.
    i keep getting different ans, dunno where i am going wrong
  2. jcsd
  3. Oct 6, 2009 #2


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    We can't tell you what you are doing wrong if you don't tell us what you are doing!

    Also, what do you mean by X(0)? Are you thinking of X as a function of y? That is, X(0)= 100 means that if y= 0, X= 100?
  4. Oct 6, 2009 #3
    Assuming this equation is more correctly written as:
    Log(e^X) = 2 * Log[e^(100-Y^2)] + C

    and, assuming
    then we have:
    Log(e^100) = 2 * Log(e^100) + C
    and C = -Log(e^100)

    But, if you really mean to use natural log (Ln), rather than the base 10 logarithm (Log), the equation becomes:
    Ln(e^100) = 2 * Ln(e^100) + C

    and, since Ln(e^x) = x, we have:
    100 = 2 * 100 + C
    100 = 200 + C
    C = -100
  5. Oct 7, 2009 #4


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    I assumed that he does not mean log(ex) but that "Loge" meant "log base e" or natural logarithm.

    Hopefully, imperiale will come back and explain in more detail exactly what he is trying to do.
  6. Oct 7, 2009 #5
    If this is the case, then using Log base e (better known as natural logarithm, or "Ln"), the equation becomes:
    Ln(X) = 2 * Ln(100 - Y^2) + C

    and, still assuming
    we would have:
    Ln(100) = 2 * Ln(100) + C
    and C = -Ln(100)
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