Answer: \(y=\mathrm{e}^{9-2x^3}\), and \(x=-2\).

When \(\ln y\) is plotted against \(x^3\), the vertical variable is \(\ln y\) and the horizontal variable is \(x^3\). Therefore the straight line has the form

\(\ln y=mx^3+c.\)

The line passes through \((2,5)\) and \((-8,25)\), so its gradient is

\(m=\dfrac{25-5}{-8-2}=\dfrac{20}{-10}=-2.\)

Using the point \((2,5)\),

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

Hence

\(c=9.\)

So

\(\ln y=9-2x^3.\)

Exponentiating both sides gives

\(y=\mathrm{e}^{9-2x^3}.\)

For \(y=\mathrm{e}^{25}\), compare exponents:

\(25=9-2x^3.\)

So

\(-2x^3=16,\)

and hence

\(x^3=-8.\)

Therefore

\(x=-2.\)