Answer: \(y=\frac15\ln(0.6x^3+5.92)\); \(x\gt -2.14\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Since \(e^{5y}\) is plotted against \(x^3\), the straight line has equation

\(e^{5y}=mx^3+c\).

Using the two given points,

\(m=\frac{9.84-4.38}{6.54-(-2.56)}=\frac{5.46}{9.10}=0.6\).

Substitute \((-2.56,4.38)\):

\(4.38=0.6(-2.56)+c\), so \(c=5.916\), which is \(5.92\) to 3 significant figures.

Therefore

\(e^{5y}=0.6x^3+5.92\).

Taking natural logarithms gives

\(5y=\ln(0.6x^3+5.92)\), so

\(y=\frac15\ln(0.6x^3+5.92)\).

For \(y\) to exist, the logarithm argument must be positive:

\(0.6x^3+5.92\gt 0\).

Thus \(x^3\gt -\frac{5.92}{0.6}\), giving

\(x\gt -2.14\) to 3 significant figures.

This gives the required answer: \(y=\frac15\ln(0.6x^3+5.92)\); \(x\gt -2.14\).