1. **State the problem:** We need to find the growth rate $g$ given the formula for $P_g$: $$P_g = A_1 \frac{1 - \left(\frac{1+g}{1+i}\right)^n}{i - g}$$ with values $P_g=7333$, $i=0.12$, $A_1=1000$, and $n=10$. 2. **Write the formula with given values:** $$7333 = 1000 \times \frac{1 - \left(\frac{1+g}{1+0.12}\right)^{10}}{0.12 - g}$$ 3. **Simplify the denominator:** $$7333 = 1000 \times \frac{1 - \left(\frac{1+g}{1.12}\right)^{10}}{0.12 - g}$$ 4. **Divide both sides by 1000:** $$7.333 = \frac{1 - \left(\frac{1+g}{1.12}\right)^{10}}{0.12 - g}$$ 5. **Multiply both sides by $(0.12 - g)$:** $$7.333 (0.12 - g) = 1 - \left(\frac{1+g}{1.12}\right)^{10}$$ 6. **Expand left side:** $$0.88 - 7.333 g = 1 - \left(\frac{1+g}{1.12}\right)^{10}$$ 7. **Rearrange to isolate the power term:** $$\left(\frac{1+g}{1.12}\right)^{10} = 1 - 0.88 + 7.333 g = 0.12 + 7.333 g$$ 8. **Take the 10th root:** $$\frac{1+g}{1.12} = \left(0.12 + 7.333 g\right)^{\frac{1}{10}}$$ 9. **Multiply both sides by 1.12:** $$1 + g = 1.12 \times \left(0.12 + 7.333 g\right)^{\frac{1}{10}}$$ 10. **This is a nonlinear equation in $g$ that can be solved numerically.** Using numerical methods (e.g., Newton-Raphson or trial and error), approximate $g \approx 0.05$ (or 5%). **Final answer:** $$g \approx 0.05$$