anonymous
  • anonymous
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!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
\[e^{x-3} = -e^{-x}\]Is that the equation? Are you doing exponentials with complex numbers?
anonymous
  • anonymous
No it is e^x - 3 = -e^-x
UnkleRhaukus
  • UnkleRhaukus
\[e^x-3=-e^{-x}\]

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UnkleRhaukus
  • UnkleRhaukus
are you trying to solve for x?
anonymous
  • anonymous
Yes! I believe that's what you do when you are solving An exponential equation right?
UnkleRhaukus
  • UnkleRhaukus
\[e^x-3=-e^{-x}\]\[e^x-3+e^{-x}=0\]\[e^{2x}-3e^x+1=0\] now substitute \(e^x=m\) \[m^2-3m+1=0\] solve this for m
anonymous
  • anonymous
Would it be sqrt 3m-1, and -sqrt 3m-1
anonymous
  • anonymous
Sorry I'm on my iPad so I. Not typing it correctly
UnkleRhaukus
  • UnkleRhaukus
i got \[m=\frac{3\pm\sqrt5}{2}\]
anonymous
  • anonymous
Woahhh, okay I'm way off! How did you come up with that?
UnkleRhaukus
  • UnkleRhaukus
quadratic formula?
UnkleRhaukus
  • UnkleRhaukus
to get x , just take the natural logarithm of m
anonymous
  • anonymous
Got it, was doing it wrong :-) thank you!!!!
UnkleRhaukus
  • UnkleRhaukus
ill check your final result when you get there, if you want
whpalmer4
  • whpalmer4
@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? :-)
UnkleRhaukus
  • UnkleRhaukus
note that: \[\cosh x=\frac 32\]\(\qquad\Downarrow\)\[x=\pm\operatorname{arccosh}\frac 32\]

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