Results 1 to 3 of 3

Thread: Condensing Logs :(

  1. April 7th 2011, 07:55 PM #1
    layneek
    layneek is offline
    Newbie
    Joined
    Apr 2011
    Posts
    1

    Condensing Logs :(

    So i have been working on this problem for about 45 minutes now and cant really seem to be going anywhere. I ended up getting different answers each time and have no idea what im doing wrong. Can you guys help please?


    Condense: 4(logx + 3log(x+2) + logy) - 2(log15 + log2 + logy) - 1/2log4 + 2/3log27-(4log 5 + 4log x + 2log y)-11log(x+2)
    Follow Math Help Forum on Facebook and Google+
  2. April 7th 2011, 09:03 PM #2
    topsquark
    topsquark is offline
    Moderator
    topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,080
    Thanks
    153
    Awards
    1
    MHF Donor
    Quote Originally Posted by layneek View Post
    So i have been working on this problem for about 45 minutes now and cant really seem to be going anywhere. I ended up getting different answers each time and have no idea what im doing wrong. Can you guys help please?


    Condense: 4(logx + 3log(x+2) + logy) - 2(log15 + log2 + logy) - 1/2log4 + 2/3log27-(4log 5 + 4log x + 2log y)-11log(x+2)
    You have two rules to follow:
    log(a) + log(b) = log(ab)

    and
    a \cdot log(b) = log(b^a)

    Your first term, for example:
    4(log(x) + 3log(x+2) + log(y)) = 4(log[x(x + 2)]y) = log \{ ( [ x(x + 2) ] y)^4 \} = log(x^4(x + 2)^4 y^4)

    Now do this with your other terms.

    -Dan
    Follow Math Help Forum on Facebook and Google+
  3. April 7th 2011, 09:25 PM #3
    Soroban
    Soroban is offline
    Super Member
    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,243
    Thanks
    370
    Hello, layneek!

    \text{Condense:}

    4\bigg[\log x + 3\log(x+2) + \log y\bigg] - 2\bigg[\log15 + \log2 + \log y\bigg]

    . . . - \frac{1}{2}\log4 + \frac{2}{3}\log27-\bigg[4\log 5 + 4\log x + 2\log y\bigg]-11\log(x+2)

    Clear parentheses:

    4\log x + 12\log(x+2) + 4\log y - 2\log15 - 2\log 2 - 2 \log y
    . . . . -\frac{1}{2}\log4 + \frac{2}{3}\log27 - 4\log5 - 4\log x - 2\log y - 11\log(x+2)

    . . =\;\log(x+2) + \frac{2}{3}\log27 -2\log15 - 2\log2 -\frac{1}{2}\log4 - 4\log5

    . . =\;\log(x+2) + \log\left(27^{\frac{2}{3}}\right) - \log\left(15^2\right) - \log(2^2) - \log\left(4^{\frac{1}{2}}\right) - \log(5^4)

    . . =\;\bigg[\log(x+2) + \log 9\bigg] - \bigg[\log225 + \log 4 + \log 2 + \log625\bigg]

    . . =\;\log9(x+2) - \log(225\cdot4\cdot2\cdot625)

    . . =\;\log9(x+2) - \log(1,\!125,\!000)

    . . =\;\log\dfrac{9(x+2)}{1,125,000}

    Follow Math Help Forum on Facebook and Google+
« 48 ÷ x(9+3) = 288 | Why is the same variable capable of expressing 2 different numbers? »

Similar Math Help Forum Discussions

  1. Condensing to One Radical
    Posted in the Algebra Forum
    Replies: 6
    Last Post: May 18th 2010, 12:32 AM
  2. Condensing Logarithms
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 16th 2010, 08:49 AM
  3. Condensing Logarithm Help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: November 18th 2009, 01:00 PM
  4. Condensing Logarithms
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 29th 2009, 10:51 AM
  5. Expanding/Condensing Logs
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 30th 2007, 08:26 PM

Search Tags


/mathhelpforum @mathhelpforum