Results 1 to 4 of 4

Thread: Elementary Functions Logarithm question

  1. #1
    Senior Member
    Joined
    Oct 2009
    From
    Detroit
    Posts
    269
    Thanks
    5

    Elementary Functions Logarithm question

    I'd like to know why log (x^2)*(y^3); is 2*log x + 3*log y, and not; 6(log x + log y).

    log (x^2)*(y^3) = 2*log x*(y^3)
    2*3*log x*y
    6*(log x + log y)

    I suppose I'm asking about the order in which I am supposed to conduct operations with logarithms. I know this process is wrong but I can not justify my wrongness and the book I am working from does not elaborate much on logarithms.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33

    Re: Elementary Functions Logarithm question

    Consider $\displaystyle \log (ab) = \log a + \log b$

    and $\displaystyle \log a^b = b\log a$

    threrefore

    $\displaystyle \displaystyle \log (x^2y^3) = \log (x^2)+ \log (y^3) = 2\log x+ 3\log y$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,771
    Thanks
    3028

    Re: Elementary Functions Logarithm question

    Quote Originally Posted by bkbowser View Post
    I'd like to know why log (x^2)*(y^3); is 2*log x + 3*log y, and not; 6(log x + log y).

    log (x^2)*(y^3) = 2*log x*(y^3)
    2*3*log x*y
    6*(log x + log y)

    I suppose I'm asking about the order in which I am supposed to conduct operations with logarithms. I know this process is wrong but I can not justify my wrongness and the book I am working from does not elaborate much on logarithms.
    $\displaystyle x^2= x*x$, $\displaystyle y^3= y*y*y$ so $\displaystyle x^2y^3$ would be x*x*y*y*y. $\displaystyle 6(log x+ log y)= 6log(xy)=log(x^6y^6)$ and $\displaystyle x^6y^6= x*x*x*x*x*x*y*y*y*y*y*y$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

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

    Re: Elementary Functions Logarithm question

    Hello, bkbowser!

    You are taking liberties with Algebra . . .


    $\displaystyle \text{I'd lik\,\!e to know why }\,\log (x^2y^3) \:=\:2\log x + 3\log y$
    . . $\displaystyle \text{and not }6(\log x + \log y).$

    $\displaystyle \log (x^2y^3) \:=\: 2\log (xy^3)$ .This is wrong!

    Suppose your answer is correct . . .

    You have: .$\displaystyle 6(\log x + \log y) \;=\; 6\log(xy) \;=\;\log(xy)^6 \;=\;\log(x^6y^6) $

    . . See?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Elementary Functions and Polynomials
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Jun 5th 2011, 01:59 AM
  2. Integral in terms of elementary functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 18th 2011, 07:58 PM
  3. Replies: 1
    Last Post: Aug 17th 2010, 04:43 AM
  4. Elementary Set Question
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: Sep 11th 2008, 02:05 AM
  5. Elementary Functions as a Solvable Group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 13th 2007, 12:19 PM

Search Tags


/mathhelpforum @mathhelpforum