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jennag

  • 2 years ago

E^x-3=-e^-x someone please help me figure out how to solve the exponential equation! I keep coming up with this being undefined!!!

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  1. whpalmer4
    • 2 years ago
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    \[e^{x-3} = -e^{-x}\]Is that the equation? Are you doing exponentials with complex numbers?

  2. jennag
    • 2 years ago
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    No it is e^x - 3 = -e^-x

  3. UnkleRhaukus
    • 2 years ago
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    \[e^x-3=-e^{-x}\]

  4. UnkleRhaukus
    • 2 years ago
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    are you trying to solve for x?

  5. jennag
    • 2 years ago
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    Yes! I believe that's what you do when you are solving An exponential equation right?

  6. UnkleRhaukus
    • 2 years ago
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    \[e^x-3=-e^{-x}\]\[e^x-3+e^{-x}=0\]\[e^{2x}-3e^x+1=0\] now substitute \(e^x=m\) \[m^2-3m+1=0\] solve this for m

  7. jennag
    • 2 years ago
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    Would it be sqrt 3m-1, and -sqrt 3m-1

  8. jennag
    • 2 years ago
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    Sorry I'm on my iPad so I. Not typing it correctly

  9. UnkleRhaukus
    • 2 years ago
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    i got \[m=\frac{3\pm\sqrt5}{2}\]

  10. jennag
    • 2 years ago
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    Woahhh, okay I'm way off! How did you come up with that?

  11. UnkleRhaukus
    • 2 years ago
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    quadratic formula?

  12. UnkleRhaukus
    • 2 years ago
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    to get x , just take the natural logarithm of m

  13. jennag
    • 2 years ago
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    Got it, was doing it wrong :-) thank you!!!!

  14. UnkleRhaukus
    • 2 years ago
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    ill check your final result when you get there, if you want

  15. whpalmer4
    • 2 years ago
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    @jennag just for your edification, there's a thing called the hyperbolic cosine function, \(\cosh x\) which happens to be \[\cosh x =\frac{e^x-e^{-x}}{2}\]Your problem could have been written as \[2 \cosh x = 3\]Looks so cute and innocent, doesn't it? :-)

  16. UnkleRhaukus
    • 2 years ago
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    note that: \[\cosh x=\frac 32\]\(\qquad\Downarrow\)\[x=\pm\operatorname{arccosh}\frac 32\]

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