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baldymcgee6 Group Title

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

  • one year ago
  • one year ago

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  1. lgbasallote Group Title
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    are you allowed to use trigonometry?

    • one year ago
  2. baldymcgee6 Group Title
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    it was a cosh function so yes

    • one year ago
  3. lgbasallote Group Title
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    is there no such thing as arc cosh?

    • one year ago
  4. baldymcgee6 Group Title
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    If there is I have not been taught it yet

    • one year ago
  5. baldymcgee6 Group Title
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    i'm sure there is though

    • one year ago
  6. baldymcgee6 Group Title
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    I found the arc cosh function on my calculator, but how do I do this by hand?

    • one year ago
  7. AccessDenied Group Title
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    I think if we make it into a quadratic equation with \(e^x\), you could make some progress here.

    • one year ago
  8. lgbasallote Group Title
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    "progress"

    • one year ago
  9. baldymcgee6 Group Title
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    so if we multiply everything by e^x to make it a quadratic

    • one year ago
  10. LolWolf Group Title
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    Well, we know the definition of \(\cosh x\) is: \[ \cosh x=\frac{e^x+e^{-x}}{2} \]So: \[ 2\cosh x=2.5\implies\\ x=\pm\ln(2) \]

    • one year ago
  11. LolWolf Group Title
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    (Where \(x\in\mathbb{R}\))

    • one year ago
  12. baldymcgee6 Group Title
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    where did the ln come from?

    • one year ago
  13. LolWolf Group Title
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    We have: \[ \cosh^{-1}z=\ln\left(z+\sqrt{(z+1)(z-1)}\right) \]So: \[ \cosh^{-1}1.25=\ln\left(1.25+\sqrt{(2.25)(.25)}\right)=\ln(2) \]

    • one year ago
  14. AccessDenied Group Title
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    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.

    • one year ago
  15. baldymcgee6 Group Title
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    k thanks

    • one year ago
  16. AccessDenied Group Title
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    You're welcome! :)

    • one year ago
  17. Hero Group Title
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    I was wondering if something like that could be done. Now I have my answer.

    • one year ago
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