Results 1 to 7 of 7

Math Help - A summation with logs

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    49

    A summation with logs

    What is the simplified numerical value of the sum:

    \log_{2} (1-\frac{1}{2}) + \log_{2} (1-\frac{1}{3}) + ... \log_{2} (1-\frac{1}{64})

    I'm not sure how to go about solving this problem.

    Would it be

    \log_{2} ((1-\frac{1}{2})(1-\frac{1}{3}) etc)

    I can see how other properties of logs would apply as well, but I don't know the best way to go about this, and I'm not sure how to correctly arrive at a final answer
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mikedwd View Post
    What is the simplified numerical value of the sum:

    \log_{2} (1-\frac{1}{2}) + \log_{2} (1-\frac{1}{3}) + ... \log_{2} (1-\frac{1}{64})

    I'm not sure how to go about solving this problem.

    Would it be

    \log_{2} ((1-\frac{1}{2})(1-\frac{1}{3}) etc)

    I can see how other properties of logs would apply as well, but I don't know the best way to go about this, and I'm not sure how to correctly arrive at a final answer
    write 1 - (1/2) as 1/2, 1 - (1/3) as 2/3, and so on

    now what do you see?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2008
    Posts
    49
    I see the pattern:

    \log_{2} (\frac{1}{2}) + \log_{2} (\frac{2}{3}) + \log_{2} (\frac{3}{4}) ... \log_{2} (\frac{63}{64})

    But I don't know how to go about the summation...I feel a bit lost

    (0-1) + (1- \log_{2} 3)...I really don't know
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2008
    Posts
    49
    wait is it

    \log_{2} (\frac{1}{64})

    alright well thanks for the tip, it helped me out

    But how should I write this conclusion in good mathematics? The way I got to the answer is just as important as the solution.

    Should I write
    (1/2)(2/3)=1/3
    (1/2)(2/3)(3/4)(4/5)=4/5
    or something like that? is that sufficient?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2008
    Posts
    49
    Is the correct answer \log_{2} (\frac{1}{64})

    which would make the sum -6

    ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, mikedwd!

    Absolutely correct!

    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mikedwd View Post
    Is the correct answer \log_{2} (\frac{1}{64})

    which would make the sum -6

    ?
    yes

    do you see how? rewriting and then putting them as a product causes a lot of things to cancel out, leaving you with \frac 1{64} = 2^{-6}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 05:39 PM
  2. summation
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 14th 2010, 02:01 AM
  3. summation
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: August 19th 2009, 11:07 AM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 06:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 08:58 PM

Search Tags


/mathhelpforum @mathhelpforum