Results 1 to 9 of 9

Math Help - Need help with a basic equation

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    34

    Need help with a basic equation

    Solve the equation e^x - e^{-x} = a where a is an arbitrary real number.

    I have got a hint, set e^x = t, but unfortunately that does not help me much.
    Last edited by Sabo; February 10th 2010 at 07:26 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2009
    From
    stgo
    Posts
    84
    i think that it must be e^x-e^{-x}=a. Am i correct?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    34
    Doh! Yes, you are right. I have edited the post.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2010
    Posts
    26
    Quote Originally Posted by Sabo View Post
    Doh! Yes, you are right. I have edited the post.
    e^x+e^{-x} = a \Leftrightarrow t+\frac{1}{t} = a \Leftrightarrow t^2-at+1=0

    Now solve the quadratic equation
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2009
    From
    stgo
    Posts
    84
    Quote Originally Posted by Sabo View Post
    Doh! Yes, you are right. I have edited the post.
    Ahaa! That equation is well know. Now, i'll give you some help:

    e^x+e^{-x}=a

    Let be e^x=t \implies e^{-x}= \dfrac{1}{t} then the equation transform to

    t+\dfrac{1}{t}=a

    Multiplying by t both sides:

    t^2+1=at

    can you finish it?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2009
    Posts
    34
    Well, I guess it would go something along the lines of

    t^2 -at + 1 = 0
    (t-\dfrac{a}{2})^2 - \dfrac{a^2+4}{4} = 0
    (t-\dfrac{a}{2})^2 = \dfrac{a^2+4}{4}
    t-\dfrac{a}{2} = +-\sqrt{\dfrac{a^2+4}{4}}
    t = \dfrac{a +- \sqrt{a^2+4}}{2}

    For some reason I don't think that's the end of it though...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2010
    Posts
    26
    now remember that e^x>0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Feb 2009
    From
    stgo
    Posts
    84
    Quote Originally Posted by Sabo View Post
    Well, I guess it would go something along the lines of

    t^2 -at + 1 = 0
    (t-\dfrac{a}{2})^2 - \dfrac{a^2+4}{4} = 0
    (t-\dfrac{a}{2})^2 = \dfrac{a^2+4}{4}
    t-\dfrac{a}{2} = +-\sqrt{\dfrac{a^2+4}{4}}
    t = \dfrac{a +- \sqrt{a^2+4}}{2}

    For some reason I don't think that's the end of it though...
    Remember that you are serching for x, no t. Remember that, if you have e^x=y \implies x=\ln(y)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2009
    Posts
    34
    Hmm...

    t = e^x = \dfrac{a+-\sqrt{a^2+4}}{2} \implies x = \ln(\dfrac{a+-\sqrt{a^2+4}}{2})

    I guess that can be considered a solution? Or? If it is, can we simplify it further?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with a basic equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: October 30th 2010, 12:36 PM
  2. Basic Differential Equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 27th 2010, 05:31 PM
  3. Basic equation assistance...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 8th 2009, 08:40 AM
  4. BASIC EQUATION
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 28th 2008, 03:18 AM
  5. ANOTHER BASIC EQUATION
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 28th 2008, 03:13 AM

Search Tags


/mathhelpforum @mathhelpforum