Results 1 to 6 of 6

Math Help - simplification of.logarithms

  1. #1
    Member
    Joined
    Nov 2012
    From
    Singapore
    Posts
    180

    simplification of.logarithms

    I was wondering of thsee simplifications are possible, log4(y) =log2^2(y) =log2(y)^2 OR logy since.log2^2 is 1. Thank you.so.much!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: simplification of.logarithms

    \log_4{y} = \frac{\log_2{y}}{\log_2{4}} = \frac{\log_2{y}}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2012
    From
    Singapore
    Posts
    180

    Re: simplification of.logarithms

    Thanks! But are my above simplifications correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615

    Re: simplification of.logarithms

    Hello, Tutu!

    Very imaginative thinking . . . sadly, it doesn't work.


    I was wondering of thsee simplifications are possible:
    . . \log_4(y) \:=\:\log_{2^2}(y) \:=\:\log_2(y^2)\:\text{ or }\:\log_2(y)\:\text{ since }\log_22 \,=\,1.

    That last part is false: . \log_2(y^2) does not equal \log_2(y).

    And the first part is false; we cannot move the "2" like that.

    Compare: . \log_4(64) \,=\,3\:\text{ and }\:\log_2(64)\,=\6
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426

    Re: simplification of.logarithms

    Quote Originally Posted by Tutu View Post
    Thanks! But are my above simplifications correct?
    I can't tell ... I don't know if you mean \log_4{y} or \log(4y)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2012
    From
    Singapore
    Posts
    180

    Re: simplification of.logarithms

    4 is the base of the logarithm. Sorry for not being detailed and thank you so much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 17th 2012, 10:51 AM
  2. simplification
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 1st 2011, 07:49 PM
  3. Simplification
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 14th 2011, 07:18 PM
  4. Help with simplification
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 2nd 2010, 03:07 PM
  5. simplification
    Posted in the Algebra Forum
    Replies: 9
    Last Post: March 12th 2008, 12:46 PM

Search Tags


/mathhelpforum @mathhelpforum