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Divide both sides by 2:
e^{0.3t} > 20
Take natural logarithm:
0.3t > \ln(20) \approx 2.9957
t > \frac{2.9957}{0.3} \approx 9.9857
The smallest integer greater than 9.9857 is \(t = 10\).
\[ P(9) = 2e^{2.7} \approx 2 \times 14.88 = 29.76 < 40 \]
\[ P(10) = 2e^{3} \approx 2 \times 20.0855 = 40.171 > 40 \]
So, the population first exceeds 40 after \(\boxed{10}\) full days.
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