Results 1 to 4 of 4

Math Help - Check for Extraneous Solutions

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    United States
    Posts
    13

    Check for Extraneous Solutions

    My class is having a hard time understanding how to do these because we don't understand the Inverse Property solution in the problem
    and example problem we are having trouble with is:
    Log base[2](x-6)=3+log base[2](x-1)

    A solution to this problem and extra advice would be nice.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Check for Extraneous Solutions

    Quote Originally Posted by paigesisco View Post
    My class is having a hard time understanding how to do these because we don't understand the Inverse Property solution in the problem
    and example problem we are having trouble with is:
    Log base[2](x-6)=3+log base[2](x-1)

    \\log_2(x-6)=3+\log_2(x-1)\\log_2\left(\frac{x-6}{x-1}\right)=3\\\left(\frac{x-6}{x-1}\right)=8
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2013
    From
    Oakland, CA
    Posts
    12

    Re: Check for Extraneous Solutions

    And then to check for extraneous solutions: plug your x value back into the original equation (which of course we all do every time anyway <grin>). If you'd wind up taking the log of a negative number, then that solution doesn't work and is an extraneous solution.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,915
    Thanks
    779

    Re: Check for Extraneous Solutions

    Hello, paigesisco!

    \text{Solve for }x\!:\;\;\log_2(x -6) \:=\:3 + \log_2(x-1)

    Note that the domain of the solution is: \color{blue}{x > 6}.
    . . Also that: \log_2(8) = 3

    We have: . \log_2(x-6) \:=\:\log_2(8) + \log_2(x-1)

    . . . . . . . . \log_2(x-6) \;=\;\log_2\big[8(x-1)\big]

    . . . . . . . . . . . . x-6 \;=\;8x-8

    . . . . . . . . . . . . . \text{-}7x \;=\;\text{-}2

    . . . . . . . . . . . . . . x \;=\;\tfrac{2}{7}

    But this is not in the domain.
    It is an extraneous root.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: September 8th 2011, 04:22 AM
  2. Extraneous solutions, absolute value
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 22nd 2011, 08:59 AM
  3. Can someone check my solutions?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 24th 2010, 05:47 PM
  4. [SOLVED] Extraneous Solutions - Logarithms
    Posted in the Algebra Forum
    Replies: 12
    Last Post: April 13th 2008, 10:41 AM
  5. Replies: 1
    Last Post: April 22nd 2007, 11:35 AM

Search Tags


/mathhelpforum @mathhelpforum