Answer: \(Q=15.7\), \(P=484\); \(1338\); 6th week.

(a) Initially, \(t=0\), so

\(P+Q=500.\)

After \(1\) week,

\(P+Qe^2=600.\)

Subtract the first equation from the second:

\(Q(e^2-1)=100.\)

Hence

\(Q=\frac{100}{e^2-1}=15.65\ldots.\)

So

\(Q=15.7\)

to 3 significant figures. Then

\(P=500-Q=484.3\ldots,\)

so

\(P=484.\)

(b) After \(2\) weeks,

\(b=P+Qe^4.\)

Using the unrounded values,

\(b=484.3\ldots+15.65\ldots e^4=1338\ldots.\)

So the number of bacteria is

\(1338.\)

(c) Solve

\(P+Qe^{2t}\gt1000000.\)

Using the unrounded values,

\(e^{2t}\gt\frac{1000000-P}{Q}.\)

Taking logarithms,

\(2t\gt\ln\left(\frac{1000000-P}{Q}\right).\)

So

\(t\gt \frac12\ln\left(\frac{1000000-P}{Q}\right)=5.53\ldots.\)

The first full week in which the number is greater than \(1000000\) is therefore the

\(6\text{th week}.\)