Question 518114
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\log(x)\ <\ \log(x^2\ -\ 1)]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


To write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x^2)\ <\ \log(x^2\ -\ 1)]



Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(\omega_1)\ <\ \log_b(\omega_2)\ \Leftrightarrow\ \omega_1\ <\ \omega_2]


(Because *[tex \Large \frac{d}{d\omega}\log_b(\omega)\ =\ \frac{\log_b(e)}{\omega}] is positive for all *[tex \Large \omega\ >\ 0\ \in\ \mathbb{R}] which is the domain of *[tex \Large \log_b], *[tex \Large \log_b] is increasing over its entire domain.) 


To write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ <\ x^2\ -\ 1]


Add *[tex \Large -x^2] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0\ <\ -1]


Which is absurd.  The solution set is the empty set.


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