Question

Salary You accept a new job with a starting salary of \(\$ 35,000\). You receive an \(8 \%\) raise at the start of your second year, a \(7.8 \%\) raise at the start of your third year, and a \(9.4 \%\) raise at the start of your fourth year. (a) Find your salary for the second year. (b) Find your salary for the third year. (c) Find your salary for the fourth year.

Short answer

The salary for the second year is \$37800. For the third year, it's \$40764. And for the fourth year, it's approximately \$44609.616.

Step-by-step solution

Second Year Salary

To find the second year's salary, you need to apply the 8% raise to the starting salary. The multiplication factor for this raise is \(1+0.08 = 1.08\). Multiply the starting salary by this factor to find the second year's salary: \(35000 \times 1.08 = 37800\). So the salary for the second year is \$37800.

Third Year Salary

To find the third year's salary, you need to apply the 7.8% raise to the second year's salary. The multiplication factor for this raise is \(1+0.078 = 1.078\). Multiply the second year's salary by this factor to find the third year's salary: \(37800 \times 1.078 = 40764\). So the salary for the third year is \$40764.

Fourth Year Salary

To find the fourth year's salary, you need to apply the 9.4% raise to the third year's salary. The multiplication factor for this raise is \(1+0.094 = 1.094\). Multiply the third year's salary by this factor to find the fourth year's salary: \(40764 \times 1.094 = 44609.616\). So the salary for the fourth year is \$44609.616.

Key concepts

Algebraic Expressions

Understanding algebraic expressions is crucial when tackling problems like calculating salary increases. An algebraic expression is a combination of numbers, variables, and mathematical operations (such as addition, subtraction, multiplication, and division). In the context of the salary problem, the expression starts with the initial amount and incorporates the percentage increase in the form of a decimal.

For example, to calculate an 8% raise on a starting salary, the expression is derived from the original salary multiplied by a factor that represents the increase. The original salary acts as the constant term, while the multiplication factor transforms the percent raise into a usable form — in this case, 1 + 0.08 transforms an 8% raise into a multiplicative factor. Using this formula, \( 35000 \times 1.08 \) symbolizes the algebraic expression to find the second year's salary.

Percent Change

The concept of percent change is the measurement of the degree of change over time, expressed as a percentage. This can signify an increase or a decrease and is crucial for understanding the evolution of values like salaries.

To compute percent increase, you compare the new value to the original value using the formula: \( \text{{percent change}} = \frac{{\text{{new value}} - \text{{original value}}}}{{\text{{original value}}}} \times 100% \). In this exercise, however, the percent increases are already provided, and the focus shifts to applying them correctly. To ensure precision and clarity, it's helpful to express the raises as decimals (8% as 0.08, for instance) and add 1 to perform the calculation efficiently. By sequentially applying the calculated percent increases year after year, the salary progression is accurately determined.

Exponential Growth

The salary problem provided also showcases an instance of exponential growth. In mathematics, exponential growth occurs when a quantity increases by a constant proportion—the rate—over equal periods. It's represented by the formula \( P(t) = P_0e^{rt} \), where \( P(t) \) is the amount after time \( t \), \( P_0 \) is the initial amount, \( r \) is the rate of growth, and \( e \) is the base of the natural logarithms.

Although the salary raises in the exercise don't show perfect exponential growth due to varying rates, the concept is still applicable. Here, the base salary compounds as raises are applied one after the other. Each year's salary becomes the base for the next year's calculation, leading to a compounding effect akin to exponential growth—evident as each year the salary grows by a factor of its previous amount plus an additional percentage.