The number, $v$, of a certain virus present in a sample collected by a vaccine laboratory is given by $v = me^{2t} + n$, where $m$ and $n$ are constants and $t$ is measured in days.
Initially, the number of virus present was 2000. It increased to 5000 after 1 day.

1. Form the first equation using the initial condition

When $t = 0$, $v = 2000$.
$2000 = me^{2(0)} + n$
$m + n = 2000 \quad \text{--- (1)}$

2. Form the second equation using the condition after 1 day

When $t = 1$, $v = 5000$.
$5000 = me^{2(1)} + n$
$me^2 + n = 5000 \quad \text{--- (2)}$

3. Solve the simultaneous equations for $m$

From (1), $n = 2000 - m$.
Substitute into (2):
$me^2 + (2000 - m) = 5000$
$m(e^2 - 1) = 3000$
$m = \frac{3000}{e^2 - 1} \approx 469.55 \approx 470 \text{ (3 s.f.)}$

4. Solve for $n$

$n = 2000 - \frac{3000}{e^2 - 1}$
$n \approx 1530.45 \approx 1530 \text{ (3 s.f.)}$

1. Substitute the values of $m$ and $n$ from part (a) and set $v = 1000000$

From part (a), $m = \frac{3000}{e^2 - 1} \approx 469.55$ and $n = 2000 - m \approx 1530.45$.
Set $v = 1000000$:
$469.55e^{2t} + 1530.45 = 1000000$

2. Solve for $t$ using natural logarithms

$469.55e^{2t} = 998469.55$
$e^{2t} = \frac{998469.55}{469.55} \approx 2126.43$
$2t = \ln(2126.43)$
$2t \approx 7.6622$
$t \approx 3.8311$