The given equation is \(y = Ax^n\). Taking natural logarithms on both sides, we get:

\(\\ln y = \\ln A + n \\ln x\)

This is in the form of a straight line \(y = mx + c\), where \(m = n\) and \(c = \\ln A\).

From the graph, estimate the y-intercept (\(\\ln A\)) and the gradient (\(n\)).

Estimate the y-intercept from the graph, which is approximately 0.7. Therefore, \(\\ln A = 0.7\), giving \(A = e^{0.7} \approx 2.0\).

Calculate the gradient using two points from the graph. For example, using points (0.5, 0.5) and (2.5, 1.0):

Gradient \(n = \frac{1.0 - 0.5}{2.5 - 0.5} = 0.25\).

Thus, \(A \approx 2.0\) and \(n = 0.25\).