Results 1 to 5 of 5
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By MarkFL

Math Help - log

  1. #1
    Senior Member
    Joined
    Sep 2012
    From
    Sweden
    Posts
    250
    Thanks
    6

    log

    ima solve al none negative solution för this equation ehmm idk how to write it but its ^(4) *logx*^(8)*log x = ^(16)*log x idk how to solve this
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,967
    Thanks
    1635

    Re: log

    I'm not sure what the "^4" means but I will guess that it is the base of the logarithm. I am used to that being written as a subscript, after. If that is correct then the equation is log_4(x)(log_8(x))= log_16(x).

    Now, " y= log_a(x)" is the same as " a= e^y". Do you notice that all of those bases are powers of 2? 4= 2^2, 8= 2^3 and 16= 2^4. So y= log_4(x) is the same as x= 4^y= (2^2)^y= 2^{2y} and so log_2(x)= 2y, thus y= log_{4}(2)= \frac{log_2(x)}{2}. y= log_8(x), so x= 8^y= (2^3)^y= 2^{3y}, thus 3y= log_2(x) and y= log_8(x)=  \frac{log_2(x)}{3}. Similarly, log_{16}(x)= \frac{log_2(x)}{4} so the equation can be written

    \frac{log_2(x)}{2}\frac{ln_2(x)}{3}= \frac{ln_2(x)}{16}

    If we let y= log_2(x) that is the same as \frac{y}{2}\frac{y}{3}= \frac{y^2}{6}= \frac{y}{4}.
    Can you solve that?
    Thanks from Petrus
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: log

    Do you mean:

    \log_4(x)\cdot\log_8(x)=\log_{16}(x) ?

    If so, try the identity:

    \log_{b^n}(a)= \frac{{\log_b(a)}}{n}
    Thanks from Petrus
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Sep 2012
    From
    Sweden
    Posts
    250
    Thanks
    6

    Re: log

    Quote Originally Posted by MarkFL2 View Post
    Do you mean:

    \log_4(x)\cdot\log_8(x)=\log_{16}(x) ?

    If so, try the identity:

    \log_{b^n}(a)= \frac{{\log_b(a)}}{n}
    nvm did not see halloy..
    Last edited by Petrus; November 15th 2012 at 12:31 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,869
    Thanks
    655

    Re: log

    Quote Originally Posted by Petrus View Post
    ima solve al none negative solution för this equation ehmm idk how to write it but its ^(4) *logx*^(8)*log x = ^(16)*log x idk how to solve this
    please do not use "text talk" abbreviations ... it makes the post confusing and difficult to read.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum