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*(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*(1.04)^t \\ (y)/(7.05)=(7.05)/(7.05)*(1.04)^t \\ (y)/(7.05)=1*(1.04)^t \\ (y)/(7.05)=(1.04)^t \\ \log ((y)/(7.05))=\log (1.04^t) \\ \log ((y)/(7.05))=t*\log (1.04^{}) \\ t=\frac{\log((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 ((10)/(7.05))}{\log (1.04^{})}=8.9`

Then, the