\(Given the equation xⁿy² = C, take the natural logarithm of both sides:\)

\(n \ln x + 2 \ln y = \ln C\)

The equation of the line is \(y = mx + c\), where \(m\) is the gradient.

Calculate the gradient \(m\) using the points (0.31, 1.21) and (1.06, 0.91):

\(m = \frac{0.91 - 1.21}{1.06 - 0.31} = \frac{-0.3}{0.75} = -0.4\)

Since \(m = \frac{-1}{2}n\), equate and solve for \(n\):

\(-0.4 = \frac{-1}{2}n\)

\(n = 0.8\)

Substitute \(n = 0.8\) back into the equation:

\(0.8 \ln x + 2 \ln y = \ln C\)

Using the point (0.31, 1.21):

\(0.8 \times 0.31 + 2 \times 1.21 = \ln C\)

\(0.248 + 2.42 = \ln C\)

\(\ln C = 2.668\)

Calculate \(C\):

\(C = e^{2.668} \approx 14.41\)