Results 1 to 2 of 2

Math Help - Logarithms

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    Spore
    Posts
    21

    Logarithms

    What's wrong with the workings in the attachment? Why can't I factorise?
    Attached Thumbnails Attached Thumbnails Logarithms-image.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,711
    Thanks
    632

    Re: Logarithms

    Hello, Thorpelizts!

    One error . . .


    \log_2(x+2) - \log_{\sqrt{2}}(x-1) \;=\;1 \;\hdots\;\text{Note that }x > 1.

    \log_2(x+2) - \frac{\log_2(x-1)}{\underbrace{\log_2(\sqrt{2})}_{\text{This is }\frac{1}{2}}} \;=\;1

    \log_2(x+2) - 2\log_2(x-1) \;=\;1 \quad\Rightarrow\quad \log_2(x+2) - \log_2(x-1)^2 \;=\;1

    \log_2\frac{x+2}{(x-1)^2} \;=\;1 \quad\Rightarrow\quad \frac{x+2}{(x-1)^2} \;=\;2 \quad\Rightarro\quad x+2 \;=\;2(x-1)^2

    . . x + 2 \;=\;2(x^2 - 2x + 1) \quad\Rightarrow\quad x+2 \;=\;2x^2 - 4x + 2

    . . 2x^2 - 5x \;=\;0 \quad\Rightarrow\quad x(2x-5) \;=\;0


    Therefore: . {\color{red}\rlap{/////}}x = 0,\;x = \tfrac{5}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 17th 2012, 10:51 AM
  2. Logarithms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 30th 2011, 06:41 AM
  3. Logarithms - help please!
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 17th 2009, 02:58 AM
  4. Logarithms please help!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: August 29th 2009, 05:24 PM
  5. Logarithms :(
    Posted in the Algebra Forum
    Replies: 8
    Last Post: July 21st 2009, 08:33 PM

Search Tags


/mathhelpforum @mathhelpforum