Results 1 to 11 of 11

Math Help - e^x + e^(-x )= 2 : How?

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    6

    e^x + e^(-x )= 2 : How?

    Hey all,

    How would one solve e^x + e^(-x) = 2

    (This is so basic, it's embarrassing :P)

    Complete mind blank >.<
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello Kailiebe

    Quote Originally Posted by Kailiebe View Post
    Hey all,

    How would one solve e^x + e^(-x) = 2

    (This is so basic, it's embarrassing :P)

    Complete mind blank >.<
    Multiply both sides by e^x.
    Then substitute in an appropriate way to get a known form ^-^
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Kailiebe View Post
    Hey all,

    How would one solve e^x + e^(-x) = 2

    (This is so basic, it's embarrassing :P)

    Complete mind blank >.<
    Let u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}

    So then you have u+\frac{1}{u}-2=0\Rightarrow{u^2+1-2u=0}

    Quadratic formula gives

    u=\frac{2\pm\sqrt{4-4(1)(1)}}{2}=1

    So we get

    u=1

    Back subbing we get e^x=1\Rightarrow{x=\ln(1)=0}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Mathstud28 View Post
    Let u=e^x\Rightarrow{\frac{1}{u}=e^{-x}}

    So then you have u+\frac{1}{u}-2=0\Rightarrow{u^2+1-2u=0}

    Quadratic formula gives

    u=\frac{4\pm\sqrt{4-4(1)(1)}}{2}=1

    So we get

    u=1

    Back subbing we get e^x=1\Rightarrow{ln(1)=0}
    It's better seeing that you have the form a^2-2ab+b^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Moo View Post
    It's better seeing that you have the form a^2-2ab+b^2
    I dont think so, but hey its all up to the individual, that is useful though, thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Mathstud28 View Post
    I dont think so, but hey its all up to the individual, that is useful though, thanks
    Why you ?
    I rather prefer one line to 3 lines
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Moo View Post
    Why you ?
    I rather prefer one line to 3 lines
    Why do I? Why does it matter? But if you really want to know it is because I like showing the quadratic form when teaching/posting because the factoring form is a little less apparent, and you might get people attempting to factor everything like this, which sometiems is hard/impossible, But that is just my view
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Mathstud28 View Post
    Why do I? Why does it matter? But if you really want to know it is because I like showing the quadratic form when teaching/posting because the factoring form is a little less apparent, and you might get people attempting to factor everything like this, which sometiems is hard/impossible, But that is just my view
    Well, I'm just wondering eh ?
    I think it's a good thing teaching people to get used to recognizing known forms
    That's just my view too.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,043
    Thanks
    284
    You could solve it without needing to make the substitution of u= e^x, as follows:

    <br />
e^x +e^ {-x} = 2<br />

    Multiply through by  e^x:

    <br />
e^{2x} + 1 =2 e^x <br />
    <br />
e^{2x} - 2 e^x + 1 = 0<br />
    <br />
(e^x -1)(e^x-1) = 0 <br />
    <br />
e^x = 1<br />
    <br />
x = 0<br />
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by ebaines View Post
    You could solve it without needing to make the substitution of u= e^x, as follows:

    <br />
e^x +e^ {-x} = 2<br />

    Multiply through by  e^x:

    <br />
e^{2x} + 1 =2 e^x <br />
    <br />
e^{2x} - 2 e^x + 1 = 0<br />
    <br />
(e^x -1)(e^x-1) = 0 <br />
    <br />
e^x = 1<br />
    <br />
x = 0<br />
    Yes, this is exactly what Moo was talking about in the earlier posts
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,043
    Thanks
    284
    Quote Originally Posted by Mathstud28 View Post
    Yes, this is exactly what Moo was talking about in the earlier posts
    I thought she was merely observing that you had an expression of the form  u^2 + 1 -2u=0 instead of the more familar  u^2 -2u+1 =0.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum