Results 1 to 3 of 3

Math Help - Help with natural logarithms

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    16

    Help with natural logarithms

    Can someone please tell me if I am doing this problem properly?

    e^2x-3e^x+2=0

    ln2=ln3e^x-lne^2x

    ln2=(ln3e^x)/(lne^2x)

    ln2=ln3e^-x

    ln2=-xln3e (lne cancels)

    ln2=-3x and then solving for x to get -.23
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2

    Re: Help with natural logarithms

    Your second line is wrong.
    This is how I would approach the problem:
    Let A = e^x. Then you have
    A^2 - 3A + 2 = 0
    Factor to solve for A, then convert this to a solution involving x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Help with natural logarithms



    Let,









    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Natural logarithms
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 8th 2010, 07:37 AM
  2. Natural logarithms
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: December 22nd 2008, 03:15 PM
  3. Help with natural logarithms?
    Posted in the Algebra Forum
    Replies: 25
    Last Post: March 24th 2008, 06:27 PM
  4. Natural Logarithms Help Plz!
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: February 26th 2008, 02:14 PM
  5. Natural Logarithms...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 29th 2007, 04:05 AM

Search Tags


/mathhelpforum @mathhelpforum