Question 459659
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I presume you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(b^{1.6}\right)]


Use the rule for exponents on a log argument:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.6\log_b\left(b\right)]


Use the definition of logs:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b\left(b\right)\ =\ y\  \ \ \Rightarrow\ \ b^y = b]


Therefore *[tex \Large y\ =\ \log_b(b)\ =\ 1] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1.6\log_b\left(b\right)\ =\ 1.6(1)\ =\ 1.6]


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