In an experiment, 5 million cells were stored in a container. It was observed that 40% of the cells are dying every minute. The number of cells remaining, N, after t minutes, is given by $N = A e^{k t}$, where A and k are constants.

1. Determine the initial value $A$ and the number of cells remaining after 1 minute

At $t = 0$, $N = A = 5 \times 10^6$.
Since 40% of the cells die every minute, 60% remain.
At $t = 1$, $N = 0.6 \times 5 \times 10^6 = 3 \times 10^6$.

2. Substitute the values into the equation to solve for $k$

$3 \times 10^6 = 5 \times 10^6 e^{k(1)}$
$e^k = \frac{3 \times 10^6}{5 \times 10^6}$
$e^k = \frac{3}{5}$
$k = \ln \frac{3}{5}$

1. Substitute the known values into the equation

From part (a), $A = 5\,000\,000$ and $k = \ln \frac{3}{5}$.
We want to find $t$ when $N = 2000$.
$2000 = 5\,000\,000 e^{\left(\ln \frac{3}{5}\right) t}$

2. Isolate the exponential term

$e^{\left(\ln \frac{3}{5}\right) t} = \frac{2000}{5\,000\,000}$
$e^{\left(\ln \frac{3}{5}\right) t} = \frac{1}{2500}$

3. Take natural logarithm on both sides and solve for t

$\left(\ln \frac{3}{5}\right) t = \ln \left(\frac{1}{2500}\right)$
$t = \frac{\ln \left(\frac{1}{2500}\right)}{\ln \frac{3}{5}}$
$t \approx 15.313$
$t = 15.3 \text{ minutes (to 3 s.f.)}$