A virus was spreading at a chicken farm such that 2% of chickens were infected in one day. The infected chickens were culled daily. If $N$ is the total number of chickens before the start of the virus infection, and assuming that the virus continues to spread at the same rate, explain why the number of chickens expected to be alive after $n$ days is given by $(0.98)^n N$. It is known that the virus will infect $x$ % of chicken in 7 days. Calculate to the nearest 2 significant figures, the value of $x$. Given that the number of chickens expected to be alive after $n$ days can be expressed as $Ne^{kn}$, find the value of the constant $k$.

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

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