Results 1 to 6 of 6

Math Help - Logs and inequalities

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    14

    Logs and inequalities

    My study guide states that log_2x-log_3(4x-5)>1 is x>67.7721 but I have yet to rap my head around logs. If someone could reveal to me the process behind this it would greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602

    Re: Logs and inequalities

    Start by converting them to the same base...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2011
    Posts
    14

    Re: Logs and inequalities

    Would be to much trouble to ask you to show me?

    is it okay to leave it as a inequality?

    Since the base is the same, you can actually make this log (2x/(3(4x-5))) >1 or
    log (2x/(12x-15)) >1. And this simply means that (2x/12x-15)>10.

    Temporarily assuming that 12x-15 is positive (x>1.25), you can multiply each side by this and get
    2x>120x -150. Then 2x +150>120x, then 150>118 x. Then x<150/118 (but still greater than 1.25).

    Temporarily assuming that 12x-15 is negative (x<1.25), you can do the same stuff with the sign pointing the other way, 150<118x , then x>150/118, but this is a contradiction.

    So I'm sticking with 1.25<x<150/118 (this fraction is approx 1.271 btw, so a narrow value range)

    Last edited by JamesBond16; July 28th 2011 at 08:18 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Re: Logs and inequalities

    Yes, \frac{5}{4}<x<\frac{75}{59}\,, is the correct solution for \log_{10}(2x)-\log_{10}(3(4x-5))>1

    I sure that Prove It assumed that what you meant by log_2x was \log_2(x), etc. The underscore symbol, "_" , generally means that what follows is a subscript.
    Last edited by SammyS; July 28th 2011 at 10:35 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2011
    Posts
    14

    Re: Logs and inequalities

    how how do i get x>67.7721 from this?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Re: Logs and inequalities

    Quote Originally Posted by JamesBond16 View Post
    how how do i get x>67.7721 from this?
    I tried to solve it assuming that the bases were 2 & 3 as Prove It reasonably assumed. This does give the approximate answer: x>67.7721.

    Assume: \log_{2}(x)-\log_{3}(4x-5)>1\,.

    Changing the bases gives: \frac{\ln(x)}{\ln(2)}-\frac{\ln(4x-5)}{\ln(3)}>1\,.

    Multiply both sides by ln(2) and ln(3).

    Use each side as the exponent with base e.

    Solve it graphically.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 06:39 PM
  2. Replies: 2
    Last Post: March 30th 2010, 09:46 PM
  3. Replies: 2
    Last Post: April 23rd 2009, 02:26 PM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 07:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 09:58 PM

/mathhelpforum @mathhelpforum