Answer: \(y=\ln\left(15.5-2x^3\right)\), valid for \(x\lt\sqrt[3]{\dfrac{31}{4}}\).

Let

\(X=x^3\quad\text{and}\quad Y=\mathrm{e}^y\).

The question says that plotting \(Y\) against \(X\) gives a straight line through \((1,13.5)\) and \((7.5,0.5)\).

The gradient of this straight line is

\(\dfrac{0.5-13.5}{7.5-1}=\dfrac{-13}{6.5}=-2\).

So the line has equation

\(Y=-2X+c\).

Using the point \((1,13.5)\),

\(13.5=-2(1)+c\),

so

\(c=15.5\).

Substituting back \(Y=\mathrm{e}^y\) and \(X=x^3\),

\(\mathrm{e}^y=15.5-2x^3\).

Taking natural logarithms gives

\(y=\ln\left(15.5-2x^3\right)\).

For this equation to be valid, the argument of the logarithm must be positive:

\(15.5-2x^3\gt0\).

Hence

\(x^3\lt7.75=\frac{31}{4}\).

Since the cube function is increasing, this is equivalent to

\(x\lt\sqrt[3]{\frac{31}{4}}\).