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 2 years ago
E^x3=e^x someone please help me figure out how to solve the exponential equation! I keep coming up with this being undefined!!!
 2 years ago
E^x3=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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whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1\[e^{x3} = e^{x}\]Is that the equation? Are you doing exponentials with complex numbers?

jennag
 2 years ago
Best ResponseYou've already chosen the best response.0No it is e^x  3 = e^x

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2are you trying to solve for x?

jennag
 2 years ago
Best ResponseYou've already chosen the best response.0Yes! I believe that's what you do when you are solving An exponential equation right?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2\[e^x3=e^{x}\]\[e^x3+e^{x}=0\]\[e^{2x}3e^x+1=0\] now substitute \(e^x=m\) \[m^23m+1=0\] solve this for m

jennag
 2 years ago
Best ResponseYou've already chosen the best response.0Would it be sqrt 3m1, and sqrt 3m1

jennag
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry I'm on my iPad so I. Not typing it correctly

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2i got \[m=\frac{3\pm\sqrt5}{2}\]

jennag
 2 years ago
Best ResponseYou've already chosen the best response.0Woahhh, okay I'm way off! How did you come up with that?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2quadratic formula?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2to get x , just take the natural logarithm of m

jennag
 2 years ago
Best ResponseYou've already chosen the best response.0Got it, was doing it wrong :) thank you!!!!

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2ill check your final result when you get there, if you want

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1@jennag just for your edification, there's a thing called the hyperbolic cosine function, \(\cosh x\) which happens to be \[\cosh x =\frac{e^xe^{x}}{2}\]Your problem could have been written as \[2 \cosh x = 3\]Looks so cute and innocent, doesn't it? :)

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.2note that: \[\cosh x=\frac 32\]\(\qquad\Downarrow\)\[x=\pm\operatorname{arccosh}\frac 32\]
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