Answer: \(A=10^6\), \(b=10^{3/7}\), and the length of time is \(\frac73\) units.

Transform the given relationship into a straight-line form. The gradient and intercept of the straight line then give the constants in the original model.

(a) Start with

\(P=Ab^T.\)

Taking logarithms to base 10 gives

\(\lg P=\lg A+\lg(b^T).\)

Using \(\lg(b^T)=T\lg b\),

\(\lg P=T\lg b+\lg A.\)

This is in the form \(Y=mT+c\), so a graph of \(\lg P\) against \(T\) is a straight line.

(b) The graph passes through \((0,6)\), so the intercept is

\(\lg A=6.\)

Hence

\(A=10^6.\)

The gradient is

\(\frac{12-6}{14-0}=\frac{6}{14}=\frac37.\)

But the gradient is \(\lg b\), so

\(\lg b=\frac37.\)

Therefore

\(b=10^{3/7}.\)

(c) Since \(100\) million is \(10^8\) and \(1000\) million is \(10^9\), we need

\(8\lt \lg P\lt 9.\)

Using

\(\lg P=6+\frac37T,\)

we solve

\(8\lt 6+\frac37T\lt 9.\)

Subtracting \(6\),

\(2\lt \frac37T\lt 3.\)

Multiplying by \(\frac73\),

\(\frac{14}{3}\lt T\lt 7.\)

The length of time is therefore

\(7-\frac{14}{3}=\frac73.\)