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Two Short Questions

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data

    The first:

    I was asked to evaluate (x^x)^(x^x) at f'(2). I tried to use logarithmic differentiation and ended up with a really messy left hand side of y's and y', while the right hand side was 1 + lnx + 1 + lnx. The right answer however is apparently 2^10(1 + ln2)(1 + 2ln2) and I'm not sure where I made my mistake.

    The Second:

    At noon, a bacteria culture has 200 bacteria. At 1 p.m., the bacteria population has grown to 800. I have to find the time where the population is 1800 assuming exponential growth.

    2. Relevant equations

    The First: The relevant equations are above.

    The Second: I know m(t) = m(0)e^kt, so I have 800 = 200e^k1 but when I take the ln of both sides and solve for k (I got ln4) I end up with 1800 = 200e^ln4(t), which doesn't give me ln3/ln2 (the purported correct answer).

    I'm sure I'm just missing something very simple in both of these questions, but I really appreciate it.

    Thanks in advance!
     
  2. jcsd
  3. Jun 18, 2011 #2

    eumyang

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    You're going to have to show your work here, and then we can see what happened. I did get the answer you stated.

    On this one I didn't get the answer you stated, so maybe there was a typo somewhere?
     
  4. Jun 18, 2011 #3
    For the first one I took the natural log twice basically: ln(ln(y)) = ln((x^x)ln((x^x)))

    and then on the right hand side I took ln ( (x^x)(lnx^x) ) and expanded it using log laws (I should note I used log laws to bring x^x down in the first step, as you probably noticed) to ln(x^x) + ln(ln(x^x))

    Then differentiating the left hand side gets pretty messy, and the right hand side is xlnx + ln(xlnx) =

    (lnx + 1) + (lnx + 1)/xlnx = ans

    so, 1/y 1/lny y' = ans

    I have no idea though how the 2^10 comes up.

    I'll post the original statement for the second one:

    At noon, a bacteria culture has 200 bacteria. At 1 p.m., the bacteria population has grown to 800. Assuming exponential growth, when will the bacteria population be 1800?
     
  5. Jun 18, 2011 #4

    eumyang

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    You're not done yet. If I understand you correctly, since you have two expressions equaling "ans", you have
    [tex]\left(\frac{1}{y}\right)\left(\frac{1}{\ln y}\right)y' = \ln x + 1 + \frac{\ln x + 1}{x\ln x}[/tex]
    which looks right to me. Now you have to solve for y'. Multiply both sides by y ln y, but on the right side, you're actually multiplying by
    [tex]\left( x^x \right)^{x^x} \cdot x^x \ln x^x[/tex]
    because that's what y ln y equals. (You may want to simplify the above before multiplying.)

    Once you do that, then you substitute in 2 for x, because you're asked to find f'(2).

    No, I'm wondering if the answer itself (ln 3 / ln 2) is a typo, assuming you got it from the back of the textbook or the solutions manual.
     
  6. Jun 18, 2011 #5
    I don't think it's likely that it's a mistake only because it's the answer to a question from a past exam. Other answers are: 1:30 pm, ln5/ln3, ln5/ln2 and ln(5/2). Quick question at a lower level: Are ln5/ln2 and ln(5/2) not the same thing by log laws? Could both consequently be written ln5 - ln2?

    I was wondering the same thing regarding e in some of these growth questions. I end up with something like (2500e)^k and so I'll take the ln, and have k ln(2500) because I vaguely understand that lne = 1 and so it cancels. What I don't understand fully I think is why it can't become ln2500 + lne, by log laws, and consequently ln2500 + 1. Otherwise it feels to me like I'm allowing ln(2500e) to become ln(2500) ln(e), which isn't a law to my understanding. I appreciate your help so far by the way.
     
  7. Jun 19, 2011 #6

    eumyang

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    No, no. Only
    [tex]\ln \left( \frac{5}{2} \right) = \ln 5 - \ln 2[/tex]
    I don't think you're getting these right. Also what is your original expression,
    (2500e)k
    or
    2500ek?
    Because they are two different expressions.

    If you meant the first, then when you take the log, you get
    [tex]\ln (2500e)^k = k(\ln 2500 + 1)[/tex]
    ... but if you mean the second, then when you take the log, you get
    [tex]\ln 2500e^k = \ln 2500 + k[/tex]
     
  8. Jun 20, 2011 #7
    That's, right, I completely forgot about that rule. I should go back and look at the proofs to really engrave it in my mind. Thanks!
     
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