1. **State the problem:** We have two salary progressions: Helen's salary is constant at $70000$ each year, and Jane's salary starts at $70000$ and grows exponentially by $3\%$ each year, modeled by $$J_n = 70000 \times 1.03^{n-1}$$ where $n$ is the year number. 2. **Understand the formulas:** - Helen's salary each year: $$H_n = 70000$$ (constant) - Jane's salary each year: $$J_n = 70000 \times 1.03^{n-1}$$ (exponential growth) 3. **To solve questions about these salaries, typical tasks include:** - Finding the salary in a specific year $n$. - Comparing salaries after $n$ years. - Finding when Jane's salary surpasses Helen's. 4. **Example: Find Jane's salary in year 5.** Use the formula: $$J_5 = 70000 \times 1.03^{5-1} = 70000 \times 1.03^4$$ Calculate: $$1.03^4 = 1.1255$$ (approx) So, $$J_5 = 70000 \times 1.1255 = 78785$$ (approx) 5. **Example: When does Jane's salary exceed Helen's?** Set: $$J_n > H_n$$ $$70000 \times 1.03^{n-1} > 70000$$ Divide both sides by 70000: $$\cancel{70000} \times 1.03^{n-1} > \cancel{70000}$$ $$1.03^{n-1} > 1$$ Since $1.03^{n-1}$ is increasing, solve: $$n-1 > 0 \implies n > 1$$ So Jane's salary surpasses Helen's starting from year 2. 6. **Summary:** - Use Helen's constant salary formula for any year. - Use Jane's exponential formula to find salary for any year. - Compare by setting inequalities and solving for $n$. This approach helps solve any related questions about these salaries.