Question 762727
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f] is even *[tex \LARGE \Leftrightarrow\ f(x)\ =\ f(-x)]


But *[tex \LARGE \log(a)\ =\ \log(b)\ \Leftrightarrow\ a\ =\ b]


So the question becomes


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ +\ x}{1\ -\ x}\ =^?\ \frac{1\ -\ x}{1\ +\ x]


Move the right side to the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ +\ x}{1\ -\ x}\ -\ \frac{1\ -\ x}{1\ +\ x}\ =^?\ 0]


Apply the LCD and collect like terms in the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4x}{1\ -\ x^2}\ =^?\ 0]


Only when *[tex \LARGE x\ =\ 0].  But certainly not for all *[tex \LARGE x].  Therefore *[tex \LARGE f] is not even.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f] is odd *[tex \LARGE \Leftrightarrow\ f(x)\ =\ -f(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(a)\ < 0\ \Leftrightarrow\ a\ <\ 1]


Hence *[tex \LARGE -\log(a)\ \neq\ \log(a)] unless *[tex \LARGE a\ =\ 1]


Therefore *[tex \LARGE f(x)\ =\ -f(x)] only if *[tex \LARGE x\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f] is not odd, either


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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