Results 1 to 6 of 6

Thread: determine an expression for log3 interms of x and y

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    34

    determine an expression for log3 interms of x and y

    no cal multiple choice
    Given Log 6 = x Log 8 = y determine an expression for log3 interms of x and y.

    a: x/3 - y
    b: x-y/3
    c: y/3-x
    d y-x/3


    Don't understand how to turn log 3 into log 6 and log8.. help

    ThankYou
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    283
    Thanks
    2
    remember log(xy) = log(x) + log(y). So
    log 6 = log (2*3) = log 2 + log 3.
    Also log 8 = log 2*2*2 = log 2 + log 2 + log 2 = 3log2.
    Can you proceed?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    34
    Sorry.. i can't proceed.
    If log2 + log3
    log2 + x

    3log2 = log2^3

    ..ah..>.<
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Start with the fact that

    $\displaystyle \log{6} = x\log{8}$

    $\displaystyle \log{(2\cdot 3)} = x\log{\left(2^3\right)}$

    $\displaystyle \log{2} + \log{3} = 3x\log{2}$

    $\displaystyle \log{3} = 3x\log{2} - \log{2}$.

    Call this Equation 1.


    Also, remember that

    $\displaystyle \log{6} = y$

    $\displaystyle \log{(2\cdot 3)} = y$

    $\displaystyle \log{2} + \log{3} = y$

    $\displaystyle \log{2} = y - \log{3}$.

    Substitute this back into Equation 1.


    $\displaystyle \log{3} = 3x(y - \log{3}) - (y - \log{3})$

    $\displaystyle \log{3} = 3xy - 3x\log{3} - y + \log{3}$

    $\displaystyle 0 = 3xy - 3x\log{3} - y$

    $\displaystyle 3x\log{3} = 3xy - y$

    $\displaystyle \log{3} = y - \frac{y}{3x}$.



    So none of your possibilities are correct...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, hovermet!

    Given: .$\displaystyle \log 6 = x,\;\log 8 = y$
    Determine an expression for $\displaystyle \log3$ in terms of $\displaystyle x$ and $\displaystyle y$

    . . $\displaystyle (a)\;\;\frac{x}{3} - y \qquad (b)\;\;x-\frac{y}{3} \qquad (c)\;\;\frac{y}{3}-x \qquad (d)\;\;y-\frac{x}{3}$

    Note that: .$\displaystyle \log 8 \,=\,y \quad\Rightarrow\quad \log(2^3) \,=\,y \quad\Rightarrow\quad 3\log 2 \,=\,y \quad\Rightarrow\quad \log 2 \,=\,\frac{y}{3}$


    So we have: .$\displaystyle \log 3 \;\;=\;\;\log\left(\frac{6}{2}\right) \;\;=\;\;\log 6 - \log 2 \;\;=\;\;x - \frac{y}{3}$



    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2008
    Posts
    34
    Thanks guys.. it seems like i forgot an AND between log 6 = x Log 8 = y ^^;
    In fact, i done the log test pretty well today with the understanding of the mistaken question Thanks^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Feb 1st 2014, 09:20 AM
  2. log9 (4) + log3(x) = 3 solve for x
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 3rd 2011, 02:20 PM
  3. X^2+xlog6+(log2)(log3)=0?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jul 14th 2010, 05:41 PM
  4. Express h interms of x
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 6th 2010, 06:03 AM
  5. Replies: 2
    Last Post: Feb 19th 2010, 07:19 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum