某养鸡场发生病毒传播，每天有 2% 的鸡被感染。被感染的鸡每天都会被扑杀。如果 $N$ 是病毒感染开始前的鸡的总数，并且假设病毒继续以相同的速率传播，请解释为什么预计在 $n$ 天后存活的鸡的数量由 $(0.98)^n N$ 给出。 已知该病毒将在 7 天内感染 $x$ % 的鸡。计算 $x$ 的值（保留至 2 位有效数字）。 已知在 $n$ 天后预计存活的鸡的数量可以表示为 $Ne^{kn}$，求常数 $k$ 的值。

1. Determine the proportion of chickens remaining alive after one day

Since $2\%$ of the chickens are infected and culled daily, the percentage of chickens remaining alive each day is $100\% - 2\% = 98\%$.
This means the number of chickens is multiplied by a factor of $0.98$ each day.

2. Generalise the pattern to $n$ days to show the given formula

After 1 day, number of chickens alive = $0.98N$.
After 2 days, number of chickens alive = $0.98(0.98N) = (0.98)^2 N$.
Following this pattern, after $n$ days, the number of chickens expected to be alive is $(0.98)^n N$.

1. Find the proportion of chickens alive after 7 days

Number of chickens alive after 7 days = $(0.98)^7 N$
$\approx 0.8681 N$

2. Calculate the percentage of infected chickens and round to 2 significant figures

Number of infected chickens = $N - 0.8681 N = 0.1319 N$
Percentage infected = $0.1319 \times 100\% = 13.19\%$
$x = 13.19 \approx 13$ (to 2 s.f.)

1. Equate the new expression to the expression from part (i)

$Ne^{kn} = N(0.98)^n$

2. Solve for $k$ by taking the natural logarithm

$e^{kn} = (0.98)^n$
$(e^k)^n = (0.98)^n$
$e^k = 0.98$
$k = \ln(0.98)$
$k \approx -0.0202027$
$k = -0.0202 \text{ (to 3 s.f.)}$