Question 568663
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*[tex \Large e] is NOT the antilogarithm.  *[tex \Large e] is the base of the natural logs, that is: *[tex \Large \ln(x)\ =\ \log_e(x)]


If one half-life has elapsed, then *[tex \Large \frac{y}{n}\ =\ \frac{1}{2}].


Since the half-life of the substance in question is 32 years, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{32k}\ =\ \frac{1}{2} ]


Take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{32k}\right)\ =\ \ln\left(\frac{1}{2}\right)]


Use


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


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32k\ln\left(e\right)\ =\ \ln\left(\frac{1}{2}\right)]


Use "the difference of the logs is the log of the quotient",


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = \log_b\left(\frac{x}{y}\right)]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32k\ln\left(e\right)\ =\ \ln(1)\ -\ \ln(2)] 


Use


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


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32k\ =\ \ln(1)\ -\ \ln(2)]


then use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(1)\ =\ 0]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 32k\ =\ -\ln(2)]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k =\ \frac{-\ln(2)}{32}\ \approx\ -0.022]


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