Question

Suppose that a new employee starts working at $7.05 per hour and receives a 4% raise each year. After time t, in years, his hourly wage is given by the equation y = $7.05(1.04)'. Find the amount of time after which he will be earning $10.00 per hour. 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

The hourly wage after t years is given by the expression:

$y=7.05\times(1.04)^t$

In this case, we need to find the amount of time after which he will be earning $10.00 per hour, then we have to solve for t from the above equation, like this:

$\begin{gathered} y=7.05\times(1.04)^t \\ \frac{y}{7.05}=\frac{7.05}{7.05}\times(1.04)^t \\ \frac{y}{7.05}=1\times(1.04)^t \\ \frac{y}{7.05}=(1.04)^t \\ \log (\frac{y}{7.05})=\log (1.04^t) \\ \log (\frac{y}{7.05})=t\times\log (1.04^{}) \\ t=\frac{\log(\frac{y}{7.05})}{\log(1.04^{})} \end{gathered}$

Then, we just have to replace $10.00 for y into the above expression and then calculate the value of t, like this:

$t=\frac{\log (\frac{10}{7.05})}{\log (1.04^{})}=8.9$

Then, the