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Math Help - [SOLVED] Log true or false?

  1. #1
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    [SOLVED] Log true or false?

    True or False?
    For positive real numbers a and b,
    \frac {\log_{10} b}a = \log_{10} b^{1/a}

    I think its false but I don't know how to show that it is.
    Can someone please tell me how to put fractions on this forum.
    Thanks!
    Last edited by yoman360; June 14th 2009 at 02:34 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yoman360 View Post
    True or False?
    For positive real numbers a and b,
    log_{10} b/a = log_{10} (b^{1/a})

    I think its false but I don't know how to show that it is.
    Can someone please tell me how to put fractions on this forum.
    Thanks!
    do you mean \frac {\log_{10} b}a = \log_{10} b^{1/a} or \log_{10} \left( \frac ba \right) = \log_{10} b^{1/a}

    the first case is true, the last is false. do you see why?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    do you mean \frac {\log_{10} b}a = \log_{10} b^{1/a} or \log_{10} \left( \frac ba \right) = \log_{10} b^{1/a}

    the first case is true, the last is false. do you see why?
    Yeah its the first one.
    Last edited by yoman360; June 14th 2009 at 02:44 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yoman360 View Post
    Yeah its the first one.
    ok, then there is just one step. recall that \log_a x^n = n \log_a x
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    ok, then there is just one step. recall that \log_a x^n = n \log_a x
    is this correct?
    \frac {\log_{10} b}a = \log_{10} b^{1/a}
    <-------------- correct?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yoman360 View Post
    is this correct?
    \frac {\log_{10} b}a = \log_{10} b^{1/a}
    <-------------- correct?
    don't know what you were trying to say for the first part, but the last three lines are correct, that's all you need to show
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    don't know what you were trying to say for the first part, but the last three lines are correct, that's all you need to show
    well using the change of base formula:
    log_{10} b = (log_{10} b)/log_{10} 10
     log_{10} 10=1 so...
    (log_{10} b)/1= log_{10} b

    how do you put fractions in this place?
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  8. #8
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    I think his/her last posting is about change of base, and also some confusion about notation.
    More and more, we see this being used: \log (x)\text{ for }\ln (x).
    Most computer algebra systems use that convention.
    Nevertheless, historically \log (x)\text{ meant }\log_{10} (x) hence confusion.

    Using the newer notation we get the classic change of base formula.
    \log _{10} (b) = \frac{{\log (b)}}{{\log (10)}}.
    In this usage it is the case that \log (10)\not=1.
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  9. #9
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    Quote Originally Posted by Plato View Post
    I think his/her last posting is about change of base, and also some confusion about notation.
    More and more, we see this being used: \log (x)\text{ for }\ln (x).
    Most computer algebra systems use that convention.
    Nevertheless, historically \log (x)\text{ meant }\log_{10} (x) hence confusion.

    Using the newer notation we get the classic change of base formula.
    \log _{10} (b) = \frac{{\log (b)}}{{\log (10)}}.
    In this usage it is the case that \log (10)\not=1.
    log_{10} 10= 1 Read it like this 10 to what power equals 10. The answer is 1.

    or

    log_{10} 10= x rewrite in exponential form

    10^x= 10

    x= 1 so log_{10} 10= 1

    Get it?
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  10. #10
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    log_{10} x = log x
    change of base formula:
    log_x y = \frac{{\log_{10} (y)}}{{\log_{10} (x)}}

    or
    log_x y = \frac{{\log (y)}}{{\log (x)}}
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  11. #11
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    Quote Originally Posted by yoman360 View Post
    Get it?
    What do you mean by that?
    I simply pointed out to you that you may be confused as to how \log notation is being used in your text material.
    Although in most modern texts and software \log (x)\text{ is used for the older }\ln (x).
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