Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

baldymcgee6

  • 3 years ago

Solve for x: e^x + e^-x = 2.5

  • This Question is Closed
  1. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you allowed to use trigonometry?

  2. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it was a cosh function so yes

  3. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is there no such thing as arc cosh?

  4. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If there is I have not been taught it yet

  5. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i'm sure there is though

  6. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I found the arc cosh function on my calculator, but how do I do this by hand?

  7. AccessDenied
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I think if we make it into a quadratic equation with \(e^x\), you could make some progress here.

  8. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    "progress"

  9. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so if we multiply everything by e^x to make it a quadratic

  10. LolWolf
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, we know the definition of \(\cosh x\) is: \[ \cosh x=\frac{e^x+e^{-x}}{2} \]So: \[ 2\cosh x=2.5\implies\\ x=\pm\ln(2) \]

  11. LolWolf
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (Where \(x\in\mathbb{R}\))

  12. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where did the ln come from?

  13. LolWolf
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    We have: \[ \cosh^{-1}z=\ln\left(z+\sqrt{(z+1)(z-1)}\right) \]So: \[ \cosh^{-1}1.25=\ln\left(1.25+\sqrt{(2.25)(.25)}\right)=\ln(2) \]

  14. AccessDenied
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Yeah, for that quadratic method, move everything to left side and multiply by \(e^x\): \((e^x)^2 - 2.5 e^x + 1 = 0\) You can use quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \[ e^x = \frac{2.5 \pm \sqrt{(2.5)^2 - 4}}{2} \\ e^x = \frac{2.5 \pm \sqrt{2.25}}{2} \\ e^x = \frac{2.5 \pm 1.5}{2} \\ e^x = \frac{2.5 + 1.5}{2} \text{ or } e^x = \frac{2.5 - 1.5}{2} \\ e^x = 2 \text{ or } e^x = \frac{1}{2} \] Then we can just take natural logarithms for the same as above answer.

  15. baldymcgee6
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    k thanks

  16. AccessDenied
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    You're welcome! :)

  17. Hero
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was wondering if something like that could be done. Now I have my answer.

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy