Results 1 to 3 of 3

Math Help - problem: (log2(x^2))^2 - log2(2x) = 2

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    estonia
    Posts
    6

    problem: (log2(x^2))^2 - log2(2x) = 2

    Hi.

    I have a problem with a task:

    (log2(x^2))^2 - log2(2x) = 2, solve for x.

    i dont see this working out to become a quadratic equation in any way and getting the left part on the log2 base so i could use the loga - logb = log a/b got me as far as log2 (x^2logx)/2x = 2, so I'm stuck.
    Please help.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480

    Re: problem: (log2(x^2))^2 - log2(2x) = 2

    Quote Originally Posted by mathpersson View Post
    Hi.

    I have a problem with a task:

    (log2(x^2))^2 - log2(2x) = 2, solve for x.

    i dont see this working out to become a quadratic equation in any way and getting the left part on the log2 base so i could use the loga - logb = log a/b got me as far as log2 (x^2logx)/2x = 2, so I'm stuck.
    Please help.

    Thank you.
    \displaystyle \begin{align*} \left[\log_2{\left(x^2\right)} \right]^2 - \log_2{\left(2x\right)} &= 2 \\ \left[2\log_2{(x)}\right]^2 - \left[ \log_2{(2)} + \log_2{(x)} \right] &= 2 \\ 4\left[\log_2{(x)}\right]^2 - 1 - \log_2{(x)} &= 2 \\ 4\left[\log_2{(x)}\right]^2 - \log_2{(x)} - 3 &= 0 \end{align*}

    Now solve the resulting quadratic...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    estonia
    Posts
    6

    Re: problem: (log2(x^2))^2 - log2(2x) = 2

    thank you so much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. how to solve log2 (x) = log2(2x-1)
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 22nd 2011, 05:17 AM
  2. X^2+xlog6+(log2)(log3)=0?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 14th 2010, 05:41 PM
  3. log2 + log(3x - 4) = 2logx
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: December 17th 2008, 12:16 AM

Search Tags


/mathhelpforum @mathhelpforum