Results 1 to 11 of 11
Like Tree2Thanks
  • 1 Post By Shakarri
  • 1 Post By Shakarri

Math Help - logarithm manipulation

  1. #1
    Junior Member
    Joined
    Sep 2013
    From
    south africa
    Posts
    27

    logarithm manipulation

    3 log base 4 x + log base x 16 = 11/2
    First I know I have to use change of base rule
    3 log base 4 x will be log x^3/log 4 =3 log x/2 log 2
    Log base x 16 will be equal to log 16/log x = 2/log x
    But this method is'nt working. How do I manipulate this log equation?
    Critism is welcome.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    584
    Thanks
    155

    Re: logarithm manipulation

    It looks like you are changing everything to log base 10. Your workings for changing base are correct but you only need to change one of them, either put 3log4(x) into base x or logx(16) into base 4
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2013
    From
    south africa
    Posts
    27

    Re: logarithm manipulation

    Can you help with that step?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    584
    Thanks
    155

    Re: logarithm manipulation

    The general rule to convert from base a to base b is log_a(y)=\frac{log_b(y)}{log_b(a)}
    So log_x(16)=\frac{log_4(16)}{log_4(x)}
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,678
    Thanks
    611

    Re: logarithm manipulation

    Hello, Bonganitedd!

    3\log_4\!x + \log_x\!16 \:=\: \tfrac{11}{2}

    I would change everything to base-2.

    . . 3\frac{\log_2\!x}{\log_2\!4} + \frac{\log_2\!16}{\log_2\!x} \:=\:\frac{11}{2} \quad\Rightarrow\quad \frac{3\log_2\!x}{2} + \frac{4}{\log_2\!x} \:=\:\frac{11}{2}


    Multiply by 2\log_2\!x\!:

    . . 3(\log_2\!x)^2 + 8 \:=\:11\log_2\!x \quad\Rightarrow\quad 3(\log_2\!x)^2 - 11\log_2\!x + 8 \:=\:0


    Factor: . (\log_2\!x - 1)(3\log_2\!x - 8) \:=\:0


    Therefore: . \begin{Bmatrix}\log_2\!x -1 \:=\:0 & \Rightarrow & \log_2\!x \:=\:1 & \Rightarrow & x \:=\:2 \\ 3\log_2\!x -8 \:=\:0 & \Rightarrow & \log_2\!x \:=\:\frac{8}{3} & \Rightarrow & x \:=\:2^{\frac{8}{3}} \end{Bmatrix}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2013
    From
    south africa
    Posts
    27

    Re: logarithm manipulation

    I wl prefer changing to base 4.@ Soroban nt rejecting your solution bt my memo changed the equation to base 4

    And it will be 3[ log base 4 (x^2)] + log base (4) 16/log 4 (x) =11/2
    Now iam stuck with how the first expression became 3[log base 4 (x^2)]
    Infact the (x^2) part
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: logarithm manipulation

    Quote Originally Posted by Bonganitedd View Post
    I wl prefer changing to base 4.@ Soroban nt rejecting your solution bt my memo changed the equation to base 4 . And it will be 3[ log base 4 (x^2)] + log base (4) 16/log 4 (x) =11/2
    Now iam stuck with how the first expression became 3[log base 4 (x^2)]
    Infact the (x^2) part
    @Bonganitedd What you have written is totally goobly-gook.
    If you cannot express yourself in reasonable standard English do not expect help at this site.

    We are glad to help persons who are serious and have a basic grasp of basics.
    Sadly, that does not seemly apply to you.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2013
    From
    south africa
    Posts
    27

    Re: logarithm manipulation

    Ok I am sory abt dat.
    What i was trying to say is, I have a marking memoranda for the above question, but I am lost@ step 2. I will rewrite it.
    3 log base(4) x + log base (x) 16 = 11/2(original log equation)
    According to my marking memoranda
    3 log base (4) x + log base (4) 16/log base (4) x =11/2. (Change the base law)
    3[log base (4) x^2] + 2 = 11/2 log base (4) ( this is where I am lost)
    Does the equation make sense now?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2013
    From
    south africa
    Posts
    27

    Re: logarithm manipulation

    I can simplify log base (4) 16/log base (4) x to become 2/log base (4) x
    Now I will have 3 log base (4) x + 2/log base (4) x = 11/2
    Then I multiply out by log base (4) x

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Jul 2012
    From
    INDIA
    Posts
    826
    Thanks
    209

    Re: logarithm manipulation

    logarithm manipulation-20-sep-13.png
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Sep 2013
    From
    south africa
    Posts
    27

    Re: logarithm manipulation

    3 log base (4) x + 2/log base (4) x =11/2 (multiply out by log base (4) x)
    Then am having 3 (log base (4) x )(log base(4) x) +2 =log 11/2 log base(4) x
    Then 3(log base (4) x)^2 + 2 = 11/2 log base (4) x (multiply by 2)
    6(log base (4) x)^2 + 4 =11 log base (4) x
    6(log base (4) x)^2 - 11 log base (4) x + 4 = 0
    (3 log base (4) x - 4)(2 log base (4) x -1) = 0
    Log base (4) x = 4/3 or log base (4) x = 1/2
    X = 4^4/3 or x = 2
    I learned in hard way.
    Much thanks to Shakarri, Plato, Soroban and ibdutt
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equation manipulation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 10th 2011, 03:11 PM
  2. Replies: 3
    Last Post: July 18th 2011, 07:04 AM
  3. Set manipulation
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 25th 2011, 11:10 PM
  4. Set Manipulation
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 30th 2009, 03:17 PM
  5. Formula Manipulation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 30th 2006, 04:03 AM

Search Tags


/mathhelpforum @mathhelpforum