\(Given the equation y² = Ae^kx, take the natural logarithm of both sides:\)

\(2 ln y = ln A + kx.\)

This is in the form of a straight line equation y = mx + c, where the slope m is k and the intercept c is ln A.

Using the points (1.5, 1.2) and (5.24, 2.7), calculate the slope k:

k = \(\frac{2.7 - 1.2}{5.24 - 1.5} \approx 0.80\).

Substitute k = 0.80 into the equation 2 ln y = ln A + 0.80x and use one of the points to find ln A:

\(At (1.5, 1.2), 2(1.2) = ln A + 0.80(1.5).\)

\(2.4 = ln A + 1.2.\)

\(ln A = 2.4 - 1.2 = 1.2.\)

A = \(e^{1.2} \approx 3.31\).