1. The problem is to find the values of $x$ for which $r(x) = 9.3$ and $r(x) = 20$ given the function $r(x) = 5 \ln(1.4)(1.4^x)$. 2. The function is given as $r(x) = 5 \ln(1.4)(1.4^x)$. We want to solve for $x$ in the equations: $$5 \ln(1.4)(1.4^x) = 9.3$$ and $$5 \ln(1.4)(1.4^x) = 20$$ 3. To isolate $1.4^x$, divide both sides by $5 \ln(1.4)$: $$\cancel{5 \ln(1.4)}(1.4^x) = \frac{9.3}{\cancel{5 \ln(1.4)}}$$ which simplifies to $$1.4^x = \frac{9.3}{5 \ln(1.4)}$$ Similarly for the second equation: $$1.4^x = \frac{20}{5 \ln(1.4)}$$ 4. Now solve for $x$ by taking the natural logarithm of both sides: $$x = \frac{\ln\left(\frac{9.3}{5 \ln(1.4)}\right)}{\ln(1.4)}$$ and $$x = \frac{\ln\left(\frac{20}{5 \ln(1.4)}\right)}{\ln(1.4)}$$ 5. Calculate the numerical values: First, compute $5 \ln(1.4)$: $$5 \ln(1.4) \approx 5 \times 0.33647 = 1.68235$$ Then, $$x_1 = \frac{\ln\left(\frac{9.3}{1.68235}\right)}{0.33647} = \frac{\ln(5.527)}{0.33647} \approx \frac{1.710}{0.33647} \approx 5.08$$ $$x_2 = \frac{\ln\left(\frac{20}{1.68235}\right)}{0.33647} = \frac{\ln(11.88)}{0.33647} \approx \frac{2.475}{0.33647} \approx 7.35$$ 6. Final answers: $$x \approx 5.08 \text{ when } r(x) = 9.3$$ $$x \approx 7.35 \text{ when } r(x) = 20$$