Answer
$A=4121e^{0.006t}$
Work Step by Step
Exponential growth and decay models are given by $A=A_{0}e^{kt}$
in which $t$ represents time,
$A_{0}$ is the amount present at $t=0$, and
$A$ is the amount present at time $t$.
If $k>0$, the model describes growth and $k$ is the growth rate.
If $k<0$, the model describes decay and $k$ is the decay rate.
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Insert given values into the formula $A=A_{0}e^{kt}$ and find k:
$A_{0}=4121, $
$ t=40,\qquad$ (the year 2050 is 40 years after 2010)
$A=5231$
$5231=4121e^{k\cdot 40}\qquad /\div 4121$
$\displaystyle \frac{5231}{4121}=e^{40k}\qquad$... apply ln( ) to both sides...
$\displaystyle \ln\frac{5231}{4121}=40k\qquad /\div 40$
$ k=\displaystyle \frac{\ln\frac{5231}{4121}}{40}\approx$0.00596266530079$\approx 0.006$
The growth model, then, is
$A=4121e^{0.006t}$
