Results 1 to 3 of 3

Math Help - Inequality

  1. #1
    Member great_math's Avatar
    Joined
    Oct 2008
    Posts
    132

    Inequality

    Prove that \log_7 10>\log_11 13
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Quote Originally Posted by great_math View Post
    Prove that \log_7 10>\log_11 13
    Hi.
    I guess you mean  \log_7 (10)>\log_{11} (13) , right?

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

    Hence

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

    \Rightarrow \frac{ln(10)*ln(11)}{ln(7)*ln(13)} > 1
    ...
    Last edited by Rapha; November 8th 2008 at 04:26 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    Quote Originally Posted by great_math View Post
    Prove that \log_7 10>\log_{\color{red}11 }\color{red}13
    First note that the function g(x)=x\ln{x} is strictly increasing for x>1. (It is in fact strictly increasing for x>e^{-1} but we take x>1 for simplicity’s sake.)

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

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

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

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

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

    In particular, f(1)>f(3)

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

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

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

    \Rightarrow\ \frac{\ln{10}}{\ln{7}}\ >\ \frac{\ln{13}}{\ln{11}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2011, 08:20 PM
  2. Replies: 3
    Last Post: December 12th 2010, 01:16 PM
  3. inequality
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: July 25th 2010, 06:11 PM
  4. Inequality help
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 8th 2010, 06:24 AM
  5. Inequality :\
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 12th 2009, 01:57 PM

Search Tags


/mathhelpforum @mathhelpforum