The question in the image involves determining the time $t$ in years for an employee’s hourly wage to reach $10.00, starting from an initial wage of $7.62 with an annual raise of 4%.

The wage after $t$ years is given by: $y = 7.62 \times (1.04)^t$ where $y$ is the hourly wage after $t$ years.

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

1. Substitute $y = 10.00$ into the equation: $10 = 7.62 \times (1.04)^t$

2. Divide both sides by 7.62 to isolate the exponential term: $\frac{10}{7.62} = (1.04)^t$ Calculating this, we get: $1.312 \approx (1.04)^t$

3. Take the natural logarithm of both sides to solve for $t$: $\ln(1.312) = t \cdot \ln(1.04)$

4. Now, divide both sides by $\ln(1.04)$: $t = \frac{\ln(1.312)}{\ln(1.04)}$

5. Calculate the values:

   • $\ln(1.312) \approx 0.272$
   • $\ln(1.04) \approx 0.0392$

   Thus, $t \approx \frac{0.272}{0.0392} \approx 6.94$

So, the employee will be earning $10.00 per hour after approximately 6.9 years (rounded to the nearest tenth of a year).

Would you like more details on any step, or have any other questions?

Related Questions

1. What would the wage be after 10 years if the raise rate was increased to 5%?
2. How many years would it take to reach $15.00 per hour with the same conditions?
3. If the initial wage were $8.00, how would that affect the time required to reach $10.00?
4. What is the effect on the result if the raise rate were compounded monthly rather than annually?
5. How would a one-time bonus at the end of the first year affect the time calculation?

Tip:

When solving for exponential growth in finance, logarithms are essential for isolating the exponent and finding the time needed to reach a target amount.