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Math Help - Proving ln(x) is isomorphic

  1. #1
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    Proving ln(x) is isomorphic

    Need help with this problem please:
    Prove or disprove: The function ln:R+ --> R taking x to its natural logarithm, lnx, is an isomorphism.
    I am completely stuck and need help please as soon as possible! Thanks!
    Last edited by mr fantastic; May 13th 2010 at 09:22 PM.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by emlevy View Post
    Need help with this problem please:
    Prove or disprove: The function ln:R+ --> R taking x to its natural logarithm, lnx, is an isomorphism.
    I am completely stuck and need help please as soon as possible! Thanks!
    What is it you are stuck on?

    You need to prove that it is a homomorphism, and that it is a bijection.

    Homomorphism is relatively easy to prove - logs take multiplication and make them into addition which is precicely what you are wanting to do here (your first R is under multiplication, you second is under addition).

    To show it is an isomorphism it is easiest to find the inverse of log (why are these now equivalent?). So...what is the inverse of taking logs?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Swlabr View Post
    To show it is an isomorphism it is easiest to find the inverse of log (why are these now equivalent?). So...what is the inverse of taking logs?
    I really don't understand this argument. You are obviously alluding to the fact that \ln(e^x)=e^{\ln(x)}=x, but this is only that easy because they are inverse functions, which obviously requires that \ln(x) is bijective to begin with. To really prove what you said you would have to take actual definitions of the exponential functions and show they're inverses.

    The below is a very hard headed and equally circuitous approach (perhaps)

    A possible alternative is to prove that \frac{x-1}{x}\leqslant\ln(x)\leqslant x-1 which can be used (in conjunction with continuity) can be used to show surjectivity. Also, it is relatively easy to see that x < y\implies \int_1^x \frac{dx}{x}< \int_1^y \frac{dy}{y}\implies \ln(x)< \ln(y) and thus \ln is a continuous strictly increasing function and thus injective.
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  4. #4
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    Actually, the question, as stated, makes no sense. You cannot just say that a given function is an "isomorphism". An "isomorphism", like a "homomorphism" is a function between two algebraic structures that "preserves" the operations. You haven't specified what operations you are talking about! The most obvious assumption would be that we are talking about the field of real numbers as a field- but then the proposition is NOT true: ln(x+y)\ne ln(x)+ ln(y) and, in any case, ln(x) is not defined for non-positive x.

    It is true that ln(xy)= ln(x)+ ln(y) so you might be thinking of ln(x) as from the group of positive real numbers with the operation of multiplication to the group of real numbers with the operation of subtraction.

    But that should have been stated explicitely in the first post.
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