Question 314728
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Since *[tex \LARGE \frac{1}{3}\ =\ 3^{-1}], we can say that *[tex \LARGE \log_b\left(\frac{1}{3}\right)\ =\ \log_b\left(3^{-1}\right)]


But we also know that


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


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(3^{-1}\right)\ =\ -\log_b\left(3\right)\ =\ -0.792]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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