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Question about logarithms

  1. Jun 13, 2006 #1
    Quick question about logarithms

    I'm only in gr.11 right now, and I was looking at a gr.12 textbook, since this isn't really homework I decided to post this here, is that ok or is this the wrong board?

    Well anyway, it asked what the range of of the function y = b ^ n would be (and then what that means for the domain of log(b) n) and also if log(b) n could ever be negative or 0.

    Could someone please help?

    Oh and an explained answer(not just a one word one) is hoped for.
     
  2. jcsd
  3. Jun 13, 2006 #2

    HallsofIvy

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    You can't say what the range is until you give the domain! Also, it is of importance whether b is less than or larger than 0 (or equal to 0!).

    I would think, from your use of "n" rather than "x", that the domain is all positive integers but you question makes more sense if the domain is all real numbers.
     
  4. Jun 13, 2006 #3
    I'm gr.11, and I'm looking in a gr.12 textbook, I read about logarithms and a few questions asked the following:

    What is the range of the function y = b ^ n? What does this tell you about the domain of log(b)n? Explain.

    Could the value of log(b)n ever be a negative number? Explain.
    Could the value of log(b)n ever be zero? Explain.

    It's not really homework, but this is straight out of a textbook so put it here. I don't really know what to try and figure out, so I really want some help with this.

    Oh, and I can't get the imaging thing to work, so the b in the brackets is subscript just to let you know.
     
  5. Jun 13, 2006 #4
    Err, I'm not sure of all the details, that's what's in the textbook. There's no answer you can give?

    Word for word here is what it says:

    What is the range of the function y = b ^ n? What does this tell you about the domain of log(b) n? Explain.

    Could the value of log(b) n ever be a negative number? Explain.
    Could the value of log(b) n ever be zero? Explain.

    The b's in the brackets are subscript by the way, I can't seem to get the imaging thing to work.
     
  6. Jun 13, 2006 #5
    I suppose it means for different values of b and n, what kind of numbers can you get out of [tex]b^n[/tex]? I suggest you experiment with this problem, for positive and negative b and n, which (with aid of a calculator).

    Are you familiar with inverse functions and the logarithm? The question involves some simple properties of inverse functions.
     
  7. Jun 13, 2006 #6

    mathwonk

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    assuming the variable is n, and b >0, then a;; values of b^n are positive, and if n can take any real value, then all positive reals can occur as valkues of b^n.


    consequently, log can nevre be negative, but it sounds as if you need a better book. try courant calculus.
     
  8. Jun 13, 2006 #7

    mathwonk

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    wqell for some reason i cannot edit my stupid answer to the previous question. i.e. i should have said log is only defined for positive numbers, but can have as values any real number. i.e. the inpuit of exp is any real and the output is positive, so vice versa, the input of log is positive numbers and the output is all reals.
     
  9. Jun 14, 2006 #8
    If you input negative or zero numbers could the output be negative or zero?
     
  10. Jun 14, 2006 #9
    "Are you familiar with inverse functions and the logarithm? The question involves some simple properties of inverse functions."

    Oh and what do you mean by properties of inverse functions?
     
  11. Jun 14, 2006 #10

    arildno

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    You cannot have a negative real number (or zero) as input in the log function and get out a real number. That is, for the real log function, its maximal domain is the set of strictly positive real numbers.
     
  12. Jun 14, 2006 #11
    Why can't you put a negative or zero number into a log function and get out a real one what variable in the formula exactly do you test that with say the? Also, how would I do something like this? b ^ log(b)x = 1, it asks to prove that the statement is true and I don't know what to with a log in the exponent.
    Once again, the b in the brackets is subscript.
     
    Last edited: Jun 14, 2006
  13. Jun 14, 2006 #12
    Someone please help.
     
  14. Jun 14, 2006 #13

    HallsofIvy

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    Please help? Have you understood what has been said so far?

    If b is positive and the domain is all real numbers, then:

    a) bn, where n is a positive integer, is just b multiplied by itself n times: a positive number.
    b) b0= 1, a positive number.
    c) b-n, where n is a positive integer so -n is a negative integer, is 1/bn and 1 over a positive real number is a positive real number.
    d) bm/n, where m is an integer and n is a positive integer (so m/n is any rational number) is [itex]\^n\sqrt{b^m}[/itex] and [itex]\^n\sqrt{ }[/itex] is defined as the positive root.
    e) if x is irrational, it's much more "technical": there exist a sequence of rational numbers, {rn} that converge to x. bx is the limit of the sequence [itex]\left{b^{r_n}\right}[/itex]. The limit of a sequence of positive numbers cannot be negative and, in this case, cannot be 0.

    The range of the function bx, for b> 0, is the set of all positive real numbers. Since "inverting" a function, in this case going from bx to logb(x), reverses domain and range, the domain of logb(x) is the set of all positive real numbers.

    PLEASE read your problems more carefully. You are NOT asked to prove that [itex]b^{log_b(x)}= 1[/itex] because it isn't true. What you are asked to prove is that [itex]b^{log_b(x)}= x[/itex], which is precisely saying that bx and logb(x) are inverse functions. HOW you would prove that depends on what definitions of logb(x) and bx you are using.
     
  15. Jun 14, 2006 #14
    Oh woops sorry, yeah I did read the question wrong. Can't b be a negative number though(unrelated to the misread question)?
     
  16. Jun 14, 2006 #15

    VietDao29

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    Ok, I'd suggest you to re-read the book. Yes, reading the book before having a lecture is a very very good way to learn new things. You can first try to read the book, and understand what it says. Then, you can try to think a little bit further (making up some of your own questions 'why?', 'how?'... and try to answer them), then try to answer the problems the books provide to understand the concept better.
    Ok, so first, let's begin with the function: ax. This function is defined for all real x's, and for a > 0. But can a be less than 0? Why? You can think about the case when x is irrational, or some rational values of x like x = 0.5, 0.25, ...
    And it's range is [tex]] 0, \ \infty [[/tex]. You can graph some functions to see.
    Now, read the book again to see if you can answer all the questions. :)
     
    Last edited: Jun 14, 2006
  17. Jun 14, 2006 #16

    VietDao29

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    Since you are still in the reals, b cannot be negative. Why? Just think about the irrational value of x (i.e some values that cannot be expressed as any fraction m / n, where m, n are integers, e.g [tex]\sqrt{2}[/tex]). Or some rational value of x like x = 0.5, x = 0.25, ...
    Is [tex]\sqrt{-2}[/tex], or [tex](-2) ^ {\sqrt{2}}[/tex] defined in the reals? :)
    You can try to re-read the book again. If you find it too hard to understand, you can give mathwonk's recommendation a try. :)
     
    Last edited: Jun 14, 2006
  18. Jun 14, 2006 #17

    dav2008

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  19. Jun 14, 2006 #18
    Ok, so b is always a real number? Is there an explanation as to why it's treated this way, or is that just how you use it in logarithms?


    "HOW you would prove that depends on what definitions of logb(x) and bx you are using."

    Definitions of them? What do you mean? There isn't just a way to go about proving it?
     
  20. Jun 14, 2006 #19
    And also, how do you know that b ^ log(b) x = x is saying that b ^ x and log(b) x are inverse functions? Not that I'm doubting this, I'm just curious as to how you go about figuring that out.

    Oh and I read about how b in the exponential functions is > 0 in the text somewhere as well so I'm not really confused about that anymore. The above problem is the only thing confusing me right now.
     
  21. Jun 14, 2006 #20
    It might make more sense for you if you looked as log_b(x) as the inverse function to exp_b(x) = b^x (where underscore means subscript). If f(x) and g(x) are inverse functions, then if y=f(x), then by definition x=g(y), and then combining those two equations you get y=f(g(y)) or x=g(f(x)). So if log_b(x) is the inverse function to exp_b(x), then it must be that x = exp_b(log_b(x)) and x = log_b(exp_b(x)).

    (You never see it written exp_b(x) though. I just made that up.)

    I suggest you read more about the logarithm, these are very basic questions that Google can answer better than we can (and with less sass).
     
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