1. Stating the problem: Given that when $e^y$ is plotted against $x^3$, the graph is a straight line passing through points $(1, 13.5)$ and $(7.5, 0.5)$. 2. Find the equation of the straight line relating $e^y$ and $x^3$. Let $X = x^3$ and $Y = e^y$. 3. Calculate the slope $m$ of the line using points $(1, 13.5)$ and $(7.5, 0.5)$: $$ m = \frac{0.5 - 13.5}{7.5 - 1} = \frac{-13}{6.5} = -2 $$ 4. Use point-slope form $Y - Y_1 = m(X - X_1)$ with point $(1, 13.5)$: $$ e^y - 13.5 = -2(x^3 - 1) $$ 5. Simplify to get: $$ e^y = -2x^3 + 2 + 13.5 = -2x^3 + 15.5 $$ 6. Taking natural logarithm of both sides (valid when $e^y > 0$, i.e., when $-2x^3 + 15.5 > 0$), we get: $$ y = \ln(-2x^3 + 15.5) $$ 7. Solve inequality for validity domain of $x$: $$ -2x^3 + 15.5 > 0 \implies 15.5 > 2x^3 \implies x^3 < \frac{15.5}{2} = 7.75 $$ 8. Therefore, the equation is valid for: $$ x^3 < 7.75 $$ which means $$ x < \sqrt[3]{7.75} $$ Final answers: (a) $$ y = \ln(-2x^3 + 15.5) $$ (b) Valid for $$ x < \sqrt[3]{7.75} $$.