Answer
a. $A=6.04e^{0.01t}$
b. in 2040.
Work Step by Step
$a.$
$A=A_{0}e^{kt }$, and we are given:
$ A_{0 }=6.04\quad$ (for t=0 in 2000)
$A=10 \quad $when t=50, in 2050,
From this we find k:
$10=6.04e^{50k}\qquad/\div 6.04$
$\displaystyle \frac{10}{6.04}=e^{50k}\qquad$ ... apply ln( ) to both sides
$\ln \displaystyle \frac{10}{6.04} =50k\qquad /\div 50$
$k= \displaystyle \frac{\ln \frac{10}{6.04}}{50}\approx$0.0100836216209$\approx 0.01$
So, our model is
$A=6.04e^{0.01t}$
b. Find t for which A= 9.
$9=6.04e^{0.01t}\qquad/\div 6.04$
$\displaystyle \frac{9}{6.04}=e^{0.01t}\qquad$ ... apply ln( ) to both sides
$\displaystyle \ln \frac{9}{6.04}=0.01t\qquad /\times 100$
$ t=100\displaystyle \ln \frac{9}{6.04}\approx$39.8820565389$\approx 40$
40 years after 2000 is the year 2040.
