Suppose that a new employee starts working at $7.58 per hour and receives a 3% raise each year. After time t, in years, his hourly wage is given by the equation y=$7.58(1.03)^1. Find the amount of time after which he will be earning $10.00 per hour.

Question

After what amount of time will the employee be earning $10.00 per hour?
____years (Round to the nearest tenth of a year as needed.)

Answer 1

To find the amount of time \( t \) after which the employee will be earning $10.00 per hour, we can set up the equation based on the formula given:

\[ y = 7.58(1.03)^t \]

We want to find \( t \) when \( y = 10.00 \):

\[ 10.00 = 7.58(1.03)^t \]

Now, we can divide both sides by 7.58:

\[ \frac{10.00}{7.58} = (1.03)^t \]

Calculating the left side:

\[ \frac{10.00}{7.58} \approx 1.3204 \]

So, we have:

\[ 1.3204 = (1.03)^t \]

Next, we take the logarithm of both sides to solve for \( t \):

\[ \log(1.3204) = \log((1.03)^t) \]

Using the property of logarithms that allows us to bring down the exponent:

\[ \log(1.3204) = t \log(1.03) \]

Now, we can solve for \( t \):

\[ t = \frac{\log(1.3204)}{\log(1.03)} \]

Calculating the logarithms:

\[ \log(1.3204) \approx 0.1193 \] \[ \log(1.03) \approx 0.01283 \]

Now we can find \( t \):

\[ t \approx \frac{0.1193}{0.01283} \approx 9.3 \]

Rounding to the nearest tenth of a year, we find:

\[ t \approx 9.3 \text{ years} \]

Thus, the employee will be earning $10.00 per hour after approximately 9.3 years.