# His job pays a starting salary of $32,000 with a 5 percentage raise at the end of each year. Give... ## Question: His job pays a starting salary of {eq}$32,000 {/eq} with a {eq}5 \% {/eq} raise at the end of each year.

Give his salaries for the first {eq}5 {/eq} years he works.

## Calculating Salary Using the Exponential Growth Function:

Knowing that the equation of an exponential growth function is {eq}f(x) = a \cdot (1+ r)^t {/eq}, where {eq}a {/eq} is the initial salary, {eq}r {/eq} is the rate, and {eq}t {/eq} is the time interval, one can find his salary values for the first five years by plugging in his current salary and percentage raise, then solving for {eq}f(x) {/eq} given {eq}t {/eq}.

(t) Years f(x) Salary
0 $32,000.00 1$33,600.00
2 $35,280.00 3$37,044.00
4 $38,896.20 5$40,841.01

{eq}\text{Given:}\\ \bullet a = 32,000.00\\ \bullet r = 5\% = 0.05\\ \bullet t = [1, 2, 3, 4, 5]\\~\\ \begin{align} f(x) &=32,000.00 \cdot (1+0.05)^0\\ &= $32,000.00 \cdot (1.05)^0\\ &=$32,000.00 \cdot 1\\ &= $32,000.00\\~\\ f(x) &=$32,000.00 \cdot (1+0.05)^1\\ &= $32,000.00 \cdot (1.05)^1\\ &=$32,000.00 \cdot 1.05\\ &= $33,600.00\\~\\ f(x) &=$32,000.00 \cdot (1+0.05)^2\\ &= $32,000.00 \cdot (1.05)^2\\ &=$32,000.00 \cdot 1.1025\\ &= $35,280.00\\~\\ f(x) &=$32,000.00 \cdot (1+0.05)^3\\ &= $32,000.00 \cdot (1.05)^3\\ &=$32,000.00 \cdot 1.157625\\ &= $37,044.00\\~\\ f(x) &=$32,000.00 \cdot (1+0.05)^4\\ &= $32,000.00 \cdot (1.05)^4\\ &=$32,000.00 \cdot 1.2155\\ &= $38,896.20\\~\\ f(x) &=$32,000.00 \cdot (1+0.05)^5\\ &= $32,000.00 \cdot (1.05)^5\\ &=$32,000.00 \cdot 1.276281563\\ &= \$40,841.01\\ \end{align}\\ {/eq}