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Math Help - growth function and prove

  1. #1
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    growth function and prove

    [IMG]file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]Can you help for the functions and prove at the attachment ?
    Attached Thumbnails Attached Thumbnails growth function and prove-soru1.jpg  
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  2. #2
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    Quote Originally Posted by isiksoy7 View Post
    [IMG]file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]Can you help for the functions and prove at the attachment ?

    This is basic logarithms stuff, I think: use that \log_ax=\frac{\ln x}{\ln a} , so

    \frac{f(n)}{g(n)}=\frac{\frac{\ln n}{\ln a}}{\frac{\ln n}{\ln b}}=\frac{\ln b}{\ln a}

    Tonio
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  3. #3
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    The key is to show that \log_an=C_1\log_bn and \log_bn=C_2\log_an for some constants C1, C2.

    Now logrithm is not such a scary beast. Personally, I remember only three properties of logarithms.

    (1) Definition: \log_an=x iff a^x=n
    (2) \log_a(x^y)=y\log_ax
    (3) \log_a(xy)=\log_ax+\log_ay.

    It's not difficult to deduce (2) and (3) from (1), so the main thing you need to remember is that logarithm is an inverse function to power: a^{\log_ax}=x and \log_a(a^x)=x.

    So, how to express \log_an through \log_bn? Let \log_an=x. By (1), a^x=n. We need to use \log_bn, so let's take \log_b of both sides: \log_b(a^x)=\log_bn. By (2), x\log_ba=\log_bn. Recalling the definition of x, we get \log_ba\log_an=\log_bn.

    It's also easy to remember this last formula. Here is a mnemonic (not mathematical) explanation: you go from b to a ( \log_ba), then from a to n ( \log_an) and the result is from b to n ( \log_bn).
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  4. #4
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    Thanks Emakarov for your helps.
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