Question 350781
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*[tex \LARGE\ \ \ \ \ \ \ \ \ \ e^x\ =\ 3\ -\ 2e^{-x}]

Put everything into the LHS:

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ e^x\ -\ 3\ +\ 2e^{-x}\ =\ 0]

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

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ e^{2x}\ -\ 3e^x\ +\ 2\ =\ 0]

Let *[tex \Large u\ =\ e^x] and substitute:

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ u^2\ -\ 3u\ +\ 2\ =\ 0]

Factor and solve for *[tex \Large u]

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ u\ =\ 2]

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ u\ =\ 1]

Verification of the last step results left as an exercise for the student.

Substitute *[tex \Large e^x\ =\ u]

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ e^x\ =\ 2]

Take the natural log of both sides:

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ \ln\left(e^x\right)\ =\ \ln(2)]

Use *[tex \Large \log_b\left(x^n\right)\ =\ n\,\cdot\,\log_b\left(x\right)] and *[tex \Large \ln(e)\ =\ 1]

*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x\ =\ \ln(2)]

Use the same process to finish determining the other root, remembering that *[tex \Large \ln(1)\ =\ 0]

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