The problem describes a scenario where an employee’s hourly wage increases by 5% each year. The wage is represented by the equation:

$y = 7.16 \cdot (1.05)^t$

where $t$ is the time in years, and the goal is to determine how long it will take for the wage to reach $10.00 per hour.

Solution:

We need to solve for $t$ when $y = 10.00$:

$10 = 7.16 \cdot (1.05)^t$

1. Divide both sides by 7.16 to isolate the exponential term: $\frac{10}{7.16} = (1.05)^t$ $1.396 = (1.05)^t$

2. Take the natural logarithm (ln) of both sides: $\ln(1.396) = \ln((1.05)^t)$

3. Apply the logarithmic property $\ln(a^b) = b \ln(a)$: $\ln(1.396) = t \cdot \ln(1.05)$

4. Solve for $t$: $t = \frac{\ln(1.396)}{\ln(1.05)}$

Using a calculator: $\ln(1.396) \approx 0.3328 \quad \text{and} \quad \ln(1.05) \approx 0.04879$ $t = \frac{0.3328}{0.04879} \approx 6.82$

Final Answer:

The employee will be earning $10.00 per hour after approximately 6.8 years (rounded to the nearest tenth).

Let me know if you would like a detailed breakdown or have additional questions! Here are related questions to explore:

1. What happens if the annual raise is increased to 7%?
2. How would the formula change if the starting wage was $8.00?
3. What is the employee's wage after 10 years?
4. How can we use this formula to calculate yearly wages for other jobs?
5. How does compounding affect other forms of growth, such as savings?

Tip: Always check for correct rounding when working with real-world applications!