Answer: \(b=1.04\), \(A=27\); \(59\) million; \(2034\).

(a) The information gives

\(40=Ab^{10}\)

and

\(45=Ab^{13}.\)

Divide the second equation by the first:

\(\frac{45}{40}=b^3.\)

So

\(b=\left(\frac98\right)^{1/3}=1.0400\ldots.\)

Hence

\(b=1.04\)

to 2 decimal places.

Using \(40=Ab^{10}\),

\(A=\frac{40}{b^{10}}.\)

Using the unrounded value of \(b\),

\(A=27.0\ldots,\)

so

\(A=27\)

to the nearest integer.

(b) On 1 January 2020, \(t=20\). Using \(A=27\) and \(b=1.04\),

\(P=27(1.04)^{20}=59.1\ldots.\)

So the estimated population is about

\(59\)

million.

(c) Solve

\(27(1.04)^t\gt100.\)

This gives

\((1.04)^t\gt \frac{100}{27}.\)

Taking logarithms,

\(t\gt \frac{\ln(100/27)}{\ln(1.04)}=33.4\ldots.\)

The first whole number of years after 2000 is therefore \(34\), so the first year is

\(2000+34=2034.\)