Results 1 to 5 of 5

Math Help - logarithmic equations

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    24

    logarithmic equations

    Can someone please help me out?? I'm starting to confuse myself.

    log(x-1) = 2 - log (x+1)

    okay so heres what i did:
    added log (x+1) to the other side, hence log (x-1) + log (x+1) = 2
    then i used product rule . . .log (x-1)(x+1) = 2
    now i end up getting (x+1)(x-1) = 100 after getting rid of the log with base 10, and 10^2.
    x^2 = 101

    idk this just seems wrong and it is probably, so if anyone can help me out i'd appreciate that a lot. thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hi mate

    Quote Originally Posted by rafaeli View Post
    Can someone please help me out?? I'm starting to confuse myself.

    log(x-1) = 2 - log (x+1)

    This is very simple [edit: wasn't that simple, i thought log = ln,...]

    log(x-1) = 2 - log (x+1)

    log(x-1) + log (x+1) = 2

    log([x-1][x+1]) = 2

    log(x^2 - 1) = 2

    //e^()

    e^(log(x^2-1)) = e^2

    x^2 -1 = e^2

    x^2 = e^(2) +1

    x = \sqrt{e^2 +1}

    do you understand?
    Last edited by Rapha; November 10th 2008 at 09:19 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    24
    thank you so much. i understand just about all of it. except for where the e comes in. like why use e, the natural log? and why does e^log() cancel out?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Quote Originally Posted by rafaeli View Post
    thank you so much. i understand just about all of it. except for where the e comes in. like why use e, the natural log? and why does e^log() cancel out?

    ooops, im very sorry, i thought log =: log_e

    i didnt read

    [quote]
    after getting rid of the log with base 10
    [quote]

    your solution x^2 = 101 is correct

    what i meant was

    log(x-1) = 2 - log (x+1)

    log(x-1) + log (x+1) = 2

    log([x-1][x+1]) = 2

    log(x^2 - 1) = 2

    //log_b (a) = ln(a) / ln(b)

    ln(x^2 - 1)/ln(10) = 2

    ln(x^2-1) = 2ln(10)

    //e^()

    x^2-1 = e^(2ln(10))

    x = \sqrt{e^{2ln(10)}+1} = \sqrt{101}

    Sorry, my mistake :-(
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2008
    Posts
    24
    oh its okay, thanks for your help. the original answer i got just didn't seem right, but thanks for verifying it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Logarithmic equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 20th 2011, 02:12 PM
  2. Logarithmic equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 26th 2011, 05:22 AM
  3. Logarithmic Equations
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 17th 2009, 07:56 PM
  4. Logarithmic equations
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 31st 2009, 12:27 PM
  5. Help with Logarithmic Equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 24th 2008, 05:45 PM

Search Tags


/mathhelpforum @mathhelpforum