Question 987651
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_8\left(\frac{1}{4}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x)\ =\ y\ \Leftrightarrow\ b^y\ =\ x]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8^y\ =\ \frac{1}{4}]


Since *[tex \Large 2^3\ =\ 8], then *[tex \Large 8^{\frac{1}{3}}\ =\ 2], and then, since a negative exponent is a reciprocal, *[tex \Large 8^{-\frac{1}{3}}\ =\ \frac{1}{2}], then, since one-fourth is the square of one-half,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8^{-\frac{2}{3}}\ =\ \frac{1}{4}]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_8\left(\frac{1}{4}\right)\ =\ -\frac{2}{3}]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \