Question 606711
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Just write it like you did but use an underscore following log to indicate that what comes next is a subscript, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ] log_5(1/125) means


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_5\left(\frac{1}{125}\right)]


Then, let's begin:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_5\left(\frac{1}{125}\right)\ =\ x\ -\ 1]


Use the definition of logarithms:


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


Which means, for this problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^{x\,-\,1}\ =\ \frac{1}{125}]


Note that *[tex \LARGE 125\ =\ 5^3] and recall that *[tex \LARGE a^{-n}\ =\ \frac{1}{a^n}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^{x\,-\,1}\ =\ 5^{-3}]


Recall that *[tex \LARGE a^n\ =\ a^m\ \Leftrightarrow\ n\ =\ m], hence:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 1\ =\ -\ 3]


Solve for *[tex \LARGE x]


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