A vaccine was introduced to a virus in an experiment. The number, N, of virus t hours after being introduced to the vaccine is given by $N = v \times 0.9^{2t}$, where $v$ is the amount of vaccine introduced.

1. Substitute the given time and number of virus into the formula and solve for $v$.

Given $N=v\times 0.9^{2t}$.
When $t=1$, $N=81000$.

$81000=v\times 0.9^{2(1)}$

$81000=v\times 0.9^2$

$81000=0.81v$

$v=\frac{81000}{0.81}=100000$

1. Express the exponential factor in the given formula using $0.81^t$, then substitute $0.81^t=k$.

Since $0.9^2=0.81$,
$0.9^{2t}=(0.9^2)^t=0.81^t=k.$
Using $v=100000$ from part (a),
$N=v\times 0.9^{2t}=100000\times k=100000k.$

1. Find the fraction of the initial number of virus remaining after 2 hours.

At the beginning, $t=0$:

$N_0=v\times 0.9^0=v.$

After 2 hours, $t=2$:

$N_2=v\times 0.9^{2(2)}=v\times 0.9^4.$

So the fraction remaining is

$\frac{N_2}{N_0}=0.9^4=0.6561.$

2. Convert the fractional decrease into a percentage of the initial number.

Fractional decrease:

$1-0.6561=0.3439.$

Percentage decrease:

$0.3439\times 100\%=34.39\%.$