Answer: \(\lg A\approx1.3\), \(\lg b\approx0.30\); \(A\approx20\), \(b\approx2\); \(x\approx6.3\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Take logarithms to transform the model into a straight-line relationship and then use the gradient and intercept.

Take logarithms to base 10:

\(\lg y=\lg A+x\lg b.\)

This has the form \(Y=c+mx\), where \(Y=\lg y\), \(c=\lg A\), and \(m=\lg b\).

The table gives approximately

Plotting these points gives an approximately straight line.

Using the line of best fit, the gradient is about \(0.30\), so

\(\lg b\approx0.30.\)

The intercept is about \(1.3\), so

\(\lg A\approx1.3.\)

Therefore

\(A\approx10^{1.3}\approx20,\quad b\approx10^{0.30}\approx2\)

to 1 significant figure.

For \(y=1500\),

\(\lg y=\lg1500\approx3.18.\)

Using \(\lg y\approx1.3+0.30x\),

\(3.18\approx1.3+0.30x.\)

So

\(x\approx\frac{3.18-1.3}{0.30}\approx6.3.\)

This completes the solution and gives the required result.