Answer: \(\ln y\approx0.03x+1.8\), so \(A=6\), \(k=0.03\), and \(x\approx35\).

We start with the main method. Take logarithms to transform the model into a straight-line relationship and then use the gradient and intercept.

For the model \(y=A\mathrm e^{kx}\), take natural logarithms:

\(\ln y=\ln A+kx.\)

So a graph of \(\ln y\) against \(x\) should be approximately a straight line.

The values are approximately

A suitable line of best fit is

\(\ln y\approx0.03x+1.8.\)

Comparing this with \(\ln y=\ln A+kx\),

\(k\approx0.03\)

and

\(\ln A\approx1.8.\)

Thus

\(A\approx\mathrm e^{1.8}=6.05\ldots\), so to 1 significant figure \(A=6\).

Also, to 1 significant figure, \(k=0.03\).

For \(y=17\), use

\(\ln17=0.03x+1.8.\)

So

\(x=\frac{\ln17-1.8}{0.03}=34.4\ldots\)

Hence \(x\approx35\), which is within the mark-scheme accepted range.

This completes the solution and gives the required result.