Question 481887
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ +\ 5e^{-x}\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5e^{-x}\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{-x}\ =\ 1]


Multiply both sides by *[tex \Large e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ =\ e^{x}]


Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^0\ =\ 1\ \forall\ a\ \in\ \mathbb{R}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{x}\ =\ 1\ \Leftrightarrow\ x\ =\ 0]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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