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\[e^{x-3} = -e^{-x}\]Is that the equation? Are you doing exponentials with complex numbers?

No it is e^x - 3 = -e^-x

\[e^x-3=-e^{-x}\]

are you trying to solve for x?

Yes! I believe that's what you do when you are solving An exponential equation right?

Would it be sqrt 3m-1, and -sqrt 3m-1

Sorry I'm on my iPad so I. Not typing it correctly

i got \[m=\frac{3\pm\sqrt5}{2}\]

Woahhh, okay I'm way off! How did you come up with that?

quadratic formula?

to get x , just take the natural logarithm of m

Got it, was doing it wrong :-) thank you!!!!

ill check your final result when you get there, if you want

note that:
\[\cosh x=\frac 32\]\(\qquad\Downarrow\)\[x=\pm\operatorname{arccosh}\frac 32\]