Results 1 to 4 of 4

Math Help - Log Questions

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    22

    Exclamation Log Questions

    Help! Please!

    Express in the simplest form -

    log(lower)3(/lower) 23 + 1

    1/2 + 3log(lower)10(/lower) x^2

    Solve for X

    log (lower)x+1(/lower) 27 = 3

    -log(lower)3x-1(/lower)1/32 = 5
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2007
    Posts
    237
    Quote Originally Posted by mibamars View Post
    Help! Please!

    Express in the simplest form -

    log(lower)3(/lower) 23 + 1

    1/2 + 3log(lower)10(/lower) x^2

    Solve for X

    log (lower)x+1(/lower) 27 = 3

    -log(lower)3x-1(/lower)1/32 = 5
    Unreadable questions
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by mibamars View Post
    Help! Please!

    ...
    Solve for X

    log (lower)x+1(/lower) 27 = 3

    -log(lower)3x-1(/lower)1/32 = 5
    Hello,

    I assume that you mean:

    \log_{x+1}(27) = 3 \Longleftrightarrow 27 = (x+1)^3 \Longleftrightarrow 3^3 = (x+1)^3 \Longleftrightarrow 3 = x+1 \Longleftrightarrow x = 2

    -\log_{3x-1}\left( \frac{1}{32}\right) = 5 \Longleftrightarrow  \frac{1}{32} = (3x-1)^{-5}  \Longleftrightarrow 2^{-5}= (3x-1)^{-5} \Longleftrightarrow 2 = 3x-1 \Longleftrightarrow x = 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by mibamars View Post
    Help! Please!

    Express in the simplest form -

    log(lower)3(/lower) 23 + 1

    1/2 + 3log(lower)10(/lower) x^2

    ...
    Hello,

    not certain if you are looking for this:

    \log_{3}(23) + 1 \Longrightarrow \log_{3}(23) + \log_{3}(3) = \log_{3}(23 \cdot 3) = \log_{3}(69)

    \frac{1}{2}+3 \cdot \log_{10}(x^2) = \log_{10}(\sqrt{10}) +  \log_{10}(x^6) = \log_{10}( \sqrt{10} \cdot x^6)
    Last edited by earboth; July 3rd 2007 at 04:50 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum