Question 1162868
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If 2002 is year 0, then 2012 is year 10.


We need a function of the form *[tex \Large C(t)\ =\ ae^{kt}]


Where *[tex \Large a] is the year zero value (because *[tex \Large e^0\ =\ 1]) and *[tex \Large k] is the growth rate per year since 2002, and *[tex \Large t] is the year since 2002.


We are given that *[tex \Large C(0)\ =\ 4766], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(0)\ =\ 4766e^{k(0)}]


Then we are given that *[tex \Large C(10)\ =\ 46911], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4766e^{10k}\ =\ 46911]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{10k}\ =\ \frac{46911}{4766}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10k\ =\ \ln\(\frac{46911}{4766}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ \frac{\ln(46911)\,-\,\ln(4766)}{10}]


Do the arithmetic indicated to find the approximate value of *[tex \Large k], then use that value in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4766e^{kt}\ =\ 120000]


and solve for *[tex \Large t]


Sometime in the first couple of weeks of February 2016.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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