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What is the domain of y = log [(4-t)^(2/3)] ?

  1. Nov 18, 2013 #1
    Hello to all:

    I was trying to solve the following question: what is the domain of the function y = log [(4-t)^(2/3)] for Real t. Here is my reasoning:

    The domain of the log(x) is x>0.

    I would have interpreted this question in terms of the log of the cubic root of (4-t)^2. Thus, I was thinking of the domain as the solution to the following inequality:

    (4-t)^(2/3) > 0

    The cubic root has the sign of the radicand and (4-t)^2 is positive for all t. The only restriction would be when 4-t=0 and this happens for t=4. Thus, my answer to the question is: the domain is ALL reals minus 4.

    Apparently this is not the case as the answer to the problem is t<4. Now, I am suspecting this is because they are rewriting the function as: y=log[(2/3)*(4-t)] for which indeed the answer would be t<4.

    So, I am really confused as to what is going on here. Do I always need to rewrite logarithmic functions to find their domains? Why? Please, help!!!

    Thanks guys!
     
  2. jcsd
  3. Nov 18, 2013 #2
    Hi zacc...

    The item in red should be y=(2/3)log(4-t)

    Its not about rewriting the logarithmic functions.Rather it is about fulfilling the properties of the function.

    We know that logban = nlogba .

    Now,the domain of the given function needs to satisfy all the properties of logarithms .If you take domain to be R-{0} ,then clearly, the function is not defined for all the points of domain.More specifically ,the function is not defined for t≥4 .Hence,R-{0} cannot be the domain of the given function.

    Whereas for t<4 ,the function is defined everywhere in the domain.
     
    Last edited: Nov 18, 2013
  4. Nov 18, 2013 #3

    Mark44

    Staff: Mentor

    No, it would be y = (2/3)log(4 - t), which is different from what you wrote, and has a domain of t < 4. Otherwise, your other work seems reasonable. If t = 4, (2/3)log[4 - t] and log[(4 - t)2/3] are undefined, but if t > 4, log[(4 - t)2/3] is defined, while (2/3)log(4 - t) is not.

    Have you given us the complete problem? What's your reference that says that the domain is only t < 4?
     
  5. Nov 18, 2013 #4
    Tanya and Mark: Indeed I made a mistake and I should have written y=(2/3)*log(t-4). That was my intention. Sorry about that but it was quite late at night when I was writing. Thanks for the correction.

    Hi Tanya: Thanks for the reply. I still don't see why my original reasoning is flawed. That is, (t-4)2 is positive over all reals and so is its cubic root and thus its log should be defined But, please see below.

    Mark: My only reference was asking the question to Wolframalpha: "what is the domain of the function y=log[(4-t)^(2/3)]" which returns t<4 as the answer. Requesting a plot of the function returns only the branch for t<4.

    Tanya and Mark:

    Now, I just realized that if I had asked Wolframalpha the question: what is the domain of the function y=log((t-4)^2) the answer in this case is all reals except 4. So, evidently the problem has to do with the way Wolframalpha handles the roots and maybe there is something that I am missing. If any of you have any insight it would be greatly appreciated. Thanks again.
     
  6. Nov 18, 2013 #5

    Mark44

    Staff: Mentor

    The answer you got is a result of how WA handles fractional exponents. I think this is what WA is doing:
    $$(4 - t)^{2/3} = e^{ln(4 - t)^{2/3}} = e^{2/3 * ln(4 - t)}$$
     
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