Question 584061
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A constant times a log is the log of the argument to the power of the constant, i.e. *[tex \LARGE \log_b(x^n)\ =\ n\log_b(x)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x\ -\ 2)\ +\ \log(x\ +\ 2)\ -\ \log\left(\sqrt{x^2\ +\ 4}\right)]


The sum of the logs is the log of the product, i.e.  *[tex \LARGE \log_b(x) + \log_b(y) = \log_b(xy)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left((x\ -\ 2)(x\ +\ 2)\right)\ -\ \log\left(\sqrt{x^2\ +\ 4}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x^2\ -\ 4)\ -\ \log\left(\sqrt{x^2\ +\ 4}\right)]


The difference of the logs is the log of the quotient, i.e. *[tex \LARGE \log_b(x) - \log_b(y) = \log_b\left(\frac{x}{y}\right)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{x^2\ -\ 4}{\sqrt{x^2\ +\ 4}}\right)]


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