Answer: \(y=\dfrac12\ln\left(\dfrac{x^3}{2}+4\right)\), with \(x\gt -2\).

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

Let the vertical variable on the straight-line graph be \(\mathrm e^{2y}\), and let the horizontal variable be \(x^3\).

Since the graph is a straight line,

\(\mathrm e^{2y}=mx^3+c.\)

The line passes through \((2,5)\) and \((6.4,7.2)\), so its gradient is

\(m=\frac{7.2-5}{6.4-2}=\frac{2.2}{4.4}=\frac12.\)

Use the point \((2,5)\):

\(5=\frac12(2)+c.\)

Thus

\(c=4.\)

So

\(\mathrm e^{2y}=\frac{x^3}{2}+4.\)

Take natural logarithms:

\(2y=\ln\left(\frac{x^3}{2}+4\right).\)

Hence

\(y=\frac12\ln\left(\frac{x^3}{2}+4\right).\)

For \(y\) to exist as a real value, the logarithm argument must be positive:

\(\frac{x^3}{2}+4\gt 0.\)

Therefore

\(x^3\gt -8\), so

\(x\gt -2.\)

This completes the solution and gives the required result.