Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

baldymcgee6

  • 2 years ago

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

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

    are you allowed to use trigonometry?

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

    it was a cosh function so yes

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

    is there no such thing as arc cosh?

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

    i'm sure there is though

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

    "progress"

  9. baldymcgee6
    • 2 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
    • 2 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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

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

    where did the ln come from?

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

    k thanks

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

    You're welcome! :)

  17. Hero
    • 2 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

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.