Results 1 to 7 of 7

Math Help - Logarithmic Equation

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    Asia
    Posts
    17

    Logarithmic Equation

    2[\log_{2}(x+5)]^2 + \log_{2}(x-5)\log_{2}(x+5) = [\log_{4}(x^2-10+25)]^2

    I was able to simplify until it became

    2[\log_{2}(x+5)]^2 + \log_{2}(x-5)\log_{2}(x+5) = [\frac{\log_{2}(x-5)^2}{2}]^2

    Then I said: Let a= \log_{2}(x+5) and let b= \log_{2}(x-5)

    so it becomes 2a^2 + ab = (\frac{2b}{2})^2

    I don't know where to go from there. I'm trying to get a system of equation, but I can't seem to get another equation.

    Thanks in advance.
    Last edited by miguel11795; November 20th 2012 at 08:46 PM. Reason: Typo
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Logarithmic Equation

    What is the original problem in its entirety?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2012
    From
    Asia
    Posts
    17

    Re: Logarithmic Equation

    The original problem is the first equation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Logarithmic Equation

    You may factor 2a^2+ab-b^2 as (a+b)(2a-b).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2012
    From
    Asia
    Posts
    17

    Re: Logarithmic Equation

    Thank you, but I'm still unsure about where to go from there. How can we find a value for x?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Logarithmic Equation

    This gives you two equations using the zero-factor property:

    i) a+b=0

    \log_2(x+5)+\log_2(x-5)=0

    \log_2(x^2-25)=0

    x^2-25=1

    ii) 2a-b=0

    2\log_2(x+5)=\log_2(x-5)

    (x+5)^2=x-5

    Now, solve the two equations for real roots, and keep in mind we require 5<x.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2012
    From
    Asia
    Posts
    17

    Re: Logarithmic Equation

    Oh. I completely forgot about that property. Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. logarithmic equation
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 4th 2012, 01:40 PM
  2. Logarithmic equation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: August 20th 2010, 11:07 PM
  3. Logarithmic Equation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 10th 2010, 01:55 AM
  4. Logarithmic equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 18th 2008, 09:17 AM
  5. Logarithmic Equation Help
    Posted in the Business Math Forum
    Replies: 3
    Last Post: February 15th 2008, 03:32 PM

Search Tags


/mathhelpforum @mathhelpforum