Results 1 to 7 of 7

Math Help - Logarithm question

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    5

    Logarithm question

    Can someone please help me start on this:



    thanks

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by bolsius View Post
    Can someone please help me start on this:



    thanks

    Broken link

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by bolsius View Post
    Can someone please help me start on this:



    thanks

    lol.

    Can't see anything.

    But while I'm here I may as well throw down some properties of logs in case they help...

    \log (ab) = \log(a) + \log(b).

    \log(a/b) = \log(a) - \log(b) as long as b \neq 0.

    \log(e^x) = x where [tex]e[/math[ is the exponential function.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2010
    Posts
    5
    Quote Originally Posted by Deadstar View Post
    lol.

    Can't see anything.

    But while I'm here I may as well throw down some properties of logs in case they help...

    \log (ab) = \log(a) + \log(b).

    \log(a/b) = \log(a) - \log(b) as long as b \neq 0.

    \log(e^x) = x where [tex]e[/math[ is the exponential function.

    All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here

    ln (1-x) + ln (1+x) = ln (3/4)

    thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by bolsius View Post
    All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here

    ln (1-x) + ln (1+x) = ln (3/4)

    thanks
    Use the laws of logarithms to combine the two logs on the left:

    \ln(1-x)+\ln(1+x)=\ln((1-x)(1+x))=\ln(1-x^2)=\ln(3/4)

    can you take it from there?

    CB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by bolsius View Post
    All right sorry it was a copy/paste and I dont know why it didnīt show up so I write it here

    ln (1-x) + ln (1+x) = ln (3/4)

    thanks
    \ln(a) + \ln(b) = \ln(ab) (remember this rule).

    So you have...

    \ln(1-x) + \ln(1+x) = \ln((1-x)(1+x)) = \ln(1 - x^2) = \ln(3/4)

    Solve for x.

    EDIT: too slow...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2010
    Posts
    5
    Yes thanks I got it...I wasnīt sure if I had to use the laws of logarithms both side of the equal sign (right and left)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Logarithm question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 7th 2011, 09:36 AM
  2. Logarithm question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 5th 2011, 01:26 PM
  3. Logarithm question
    Posted in the Algebra Forum
    Replies: 6
    Last Post: October 11th 2009, 12:09 AM
  4. Logarithm question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 31st 2009, 09:32 AM
  5. logarithm question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 7th 2008, 09:47 PM

Search Tags


/mathhelpforum @mathhelpforum