Question 1162157
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Let *[tex \Large x] represent the number of years after 2013, hence in 2013 *[tex \Large x\ =\ 0] and in 2019, *[tex \Large x\ =\ 7]


Parent function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ ae^{bx}]


We know that when *[tex \Large x\ =\ 0], i.e. the year 2013, *[tex \Large e^{bx}\ =\ 1], therefore *[tex \Large a\ = 5.39] Billion.


And now we have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 5.39e^{bx}]


But when *[tex \Large x\ =\ 7,\ \ ]*[tex \Large  5.39e^{7b}\ =\ 10.64] billion


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{7b}\ =\ \frac{10.64}{5.39}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\(e^{7b}\)\ =\ \ln\(\frac{10.64}{5.39}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7b\ =\ \ln\(\frac{10.64}{5.39}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \frac{\ln\(\frac{10.64}{5.39}\)}{7}]


Do the arithmetic to calculate the value of *[tex \Large b] which will allow you to express the desired function and, is coincidentally the answer to the last part of your question.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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