Results 1 to 5 of 5

Math Help - Log help

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    3

    Log help

    Im missing something. Can anyone help?

    log_2 (x) + log_2 (x-4) = 2

    => log_2 (x^2-4x) = 2
    => x^2-4x = 2^2
    => where do I go from here? Ive tried various methods and cant come up with the right answer. Everything is right up to this point right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Jimbo View Post
    Im missing something. Can anyone help?

    log_2 (x) + log_2 (x-4) = 2

    => log_2 (x^2-4x) = 2
    => x^2-4x = 2^2
    => where do I go from here? Ive tried various methods and cant come up with the right answer. Everything is right up to this point right?
    You're right up to now yes. Take 4 from both sides:

    x^2 - 4x - 4 = 0 and solve using your favourite method. Remember that x > 4 because you can't take logs of 0 or negative numbers
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    3
    Thats where Im running into trouble. Ive tried several methods to solve for x and cant seem to come up right. Ive got to go for now, but Ill come back later and post some things ive tried. Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Jimbo View Post
    Thats where Im running into trouble. Ive tried several methods to solve for x and cant seem to come up right. Ive got to go for now, but Ill come back later and post some things ive tried. Thanks!
    I used the quadratic formula. For ax^2 +bx+c = 0:

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} where a = 1, b=-4 and c=-4

    Sub them in:

    x = \frac{4 \pm \sqrt{4^2-4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2}

    as \sqrt{32} = \sqrt{16}\sqrt{2} = 4\sqrt{2} we can simplify and factorise:

    \frac{4(1 \pm \sqrt{2})}{2} = 2(1 \pm \sqrt{2}).

    Since 2(1-\sqrt{2}) < 4 we discard it as a solution because log_2(x-4) is undefined.

    2(1+\sqrt{2}) > 4 so this is a solution. You can either leave it in surd form for an exact answer or round it off
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    3
    Right after I logged off I remembered the quadratic formula and was going to try it out. I always forget about that.

    Thanks for the help!
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum