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Thread: Inequality

  1. #1
    Member great_math's Avatar
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    Inequality

    Prove that $\displaystyle \log_7 10>\log_11 13$
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  2. #2
    Senior Member
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    Quote Originally Posted by great_math View Post
    Prove that $\displaystyle \log_7 10>\log_11 13$
    Hi.
    I guess you mean $\displaystyle \log_7 (10)>\log_{11} (13) $ , right?

    Do you know the formula l$\displaystyle og_a (x) = \frac{ln(x)}{ln(a)}$

    Hence

    $\displaystyle log_7 (10) = \frac{ln(10)}{ln(7)} > \frac{ln(13)}{ln(11)} = log_{11}(13)$

    $\displaystyle \Rightarrow \frac{ln(10)*ln(11)}{ln(7)*ln(13)} > 1 $
    ...
    Last edited by Rapha; Nov 8th 2008 at 04:26 AM.
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  3. #3
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by great_math View Post
    Prove that $\displaystyle \log_7 10>\log_{\color{red}11 }\color{red}13$
    First note that the function $\displaystyle g(x)=x\ln{x}$ is strictly increasing for $\displaystyle x>1.$ (It is in fact strictly increasing for $\displaystyle x>e^{-1}$ but we take $\displaystyle x>1$ for simplicity’s sake.)

    Hence for all $\displaystyle 1<a<b.$ we have $\displaystyle a\ln{a}<b\ln{b};$ i.e. $\displaystyle \frac{\ln{a}}b-\frac{\ln{b}}a<0\ \ldots\fbox{1}$

    Now define $\displaystyle f(x)=\ln(10+x)\ln(10-x)$ for $\displaystyle 0<x<9.$

    We have $\displaystyle f'(x)=\frac{\ln(10-x)}{10+x}-\frac{\ln(10+x)}{10-x}.$

    Since $\displaystyle 1<10-x<10+x$ for $\displaystyle 0<x<9,$ we have that $\displaystyle f'(x)<0$ for $\displaystyle 0<x<9$ by $\displaystyle \fbox{1}$ above.

    Thuis $\displaystyle f(x)$ is strictly decreasing for $\displaystyle 0<x<9.$

    In particular, $\displaystyle f(1)>f(3)$

    i.e. $\displaystyle \ln{11}\cdot\ln{9}\ >\ \ln{13}\cdot\ln{7}$

    But $\displaystyle \ln{10}>\ln{9}.$

    $\displaystyle \therefore\ \ln{11}\cdot\ln{10}\ >\ \ln{13}\cdot\ln{7}$

    $\displaystyle \Rightarrow\ \frac{\ln{10}}{\ln{7}}\ >\ \frac{\ln{13}}{\ln{11}}$
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