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Chapter 4: Problem 14

Wages: A worker is reviewing his pay increases over the past several years. The table below shows the hourly wage W, in dollars, that he earned as a function of time t, measured in years since the beginning of 1990. $$ \begin{array}{|c|c|} \hline \text { Time } t & \text { Wage } W \\ \hline 1 & 15.30 \\ \hline 2 & 15.60 \\ \hline 3 & 15.90 \\ \hline 4 & 16.25 \\ \hline \end{array} $$ a. By calculating ratios, show that the data in this table are exponential. (Round the quotients to two decimal places.) b. What is the yearly growth factor for the data? c. The worker can't remember what hourly wage he earned at the beginning of 1990 . Assuming that \(W\) is indeed an exponential function, determine what that hourly wage was. d. Find a formula giving an exponential model for \(W\) as a function of \(t\). e. What percentage raise did the worker receive each year?f. Given that prices increased by 34% over the decade of the 1990s, use your model to determine whether the worker鈥檚 wage increases kept pace with inflation.

Short Answer

Expert verified
The wages are exponential with a yearly growth factor of 1.02, starting at $15.00. The worker received a 2% raise each year, which wasn't enough to keep up with 34% inflation.

Step by step solution

01

Determine If Wages Are Exponential

To determine if the wage increases are exponential, calculate the ratio of successive wages:\( \frac{W(2)}{W(1)} = \frac{15.60}{15.30} \approx 1.02 \), \( \frac{W(3)}{W(2)} = \frac{15.90}{15.60} \approx 1.02 \), and \( \frac{W(4)}{W(3)} = \frac{16.25}{15.90} \approx 1.02 \). Since the ratios are consistent, the data suggests the wages are increasing exponentially.
02

Identify Yearly Growth Factor

The common ratio calculated in Step 1 reflects the yearly growth factor. It shows that each year the wages increase by a factor of approximately 1.02.
03

Calculate Initial Hourly Wage

If \(W\) is exponential, use the general formula for exponential growth \(W = W_0 \times (1.02)^t\). At \(t = 1\), \( W = 15.30 = W_0 \times (1.02)^1 \). Solving for \( W_0 \) gives \( W_0 = \frac{15.30}{1.02} \approx 15.00 \). Thus, the initial wage at the start of 1990 was approximately $15.00.
04

Establish Exponential Model Formula

From the initial wage and growth factor, create an equation: \( W = 15.00 \times (1.02)^t \), where \(W\) is the wage and \(t\) is the number of years since 1990.
05

Determine Percentage Raise Per Year

The yearly growth factor is 1.02, meaning a 2% increase per year (since 1.02 means a 2% raise).
06

Compare Wages to Inflation

Calculate the wage at the end of the 1990s using the model with \(t = 10\): \(W = 15.00 \times (1.02)^{10} \approx 18.29\). This is around a 22% increase from $15.00. Inflation increased prices by 34%, which is more than the wage increase.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wage Increase
A wage increase refers to the raise in the amount of money paid to a worker per hour over time. In the exercise, the worker's wages have increased each year since the beginning of 1990. This is evident from the wage table showing increases from $15.30 to $16.25 over four years. To assess whether these wage increments form an exponential sequence, we look at the ratio of successive wages. Calculating these ratios helps us understand that the increases are not just random but follow a specific pattern. Consistency in the ratio between each year's wage points to an exponential model of growth. Ultimately, recognizing a wage increase pattern is crucial in managing personal finance expectations and in comparing it to economic factors like inflation.
Inflation Comparison
Inflation signifies the rate at which prices for goods and services rise, diminishing purchasing power. It's necessary to compare wage increases against inflation to determine if workers' earnings are keeping pace with rising costs. In the problem, wages grew by approximately 2% per year, leading to a total increase of about 22% over ten years. During the same period, prices increased by a total of 34% due to inflation. The analysis shows that while wages did increase, they did not keep up with inflation. This suggests that, despite a nominal wage increase, the worker's real earnings鈥攔eflecting purchasing power鈥攁ctually decreased over the decade. Therefore, comparing wages to inflation is vital for understanding true income growth, as it relates to what one can afford in terms of goods and services.
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the context of wages, an exponential function can model predictable growth patterns over time. The formula used here is:\[ W = W_0 \times (1.02)^t \]Where:- \(W\) is the wage at any time \(t\),- \(W_0\) is the initial wage,- and \(1.02\) represents the growth factor.Understanding exponential growth is indispensable for making long-term projections. Unlike linear growth, which adds a fixed amount each period, exponential growth means each period鈥檚 growth builds on the prior period鈥檚 total, reflecting compound effects.Using such a model helps predict future wages and plan financial strategies accordingly.
Yearly Growth Factor
The yearly growth factor is a numerical representation showing how much a quantity increases each year. In wage analysis, this factor is crucial to understanding how wages build over time. In the problem, the wage growth factor is approximately 1.02. This means each year, the worker's wage increases by about 2% of the previous year's wage. This calculation was made by consistently finding the ratio of wages from consecutive years. This factor is important because it represents the stability of wage growth over time. It also helps in creating exponential models to predict future earnings. Knowing the growth factor allows one to make educated decisions regarding investment in skills or education, aligning wage growth with personal financial plans.

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Most popular questions from this chapter

Nearly linear or exponential data: One of the two tables below shows data that are better approximated with a linear function, and the other shows data that are better approximated with an exponential function. Make plots to identify which is which, and then use the appropriate regression to find models for both. $$ \begin{aligned} &\begin{array}{|c|c|} \hline t & f(t) \\ \hline 1 & 3.62 \\ \hline 2 & 23.01 \\ \hline 3 & 44.26 \\ \hline 4 & 62.17 \\ \hline 5 & 83.25 \\ \hline \end{array}\\\ &\begin{array}{|c|c|} \hline t & g(t) \\ \hline 1 & 3.62 \\ \hline 2 & 5.63 \\ \hline 3 & 8.83 \\ \hline 4 & 13.62 \\ \hline 5 & 21.22 \\ \hline \end{array} \end{aligned} $$

APR and APY: Recall that financial institutions sometimes report the annual interest rate that they offer on investments as the APR, often called the nominal interest rate. To indicate how an investment will actually grow, they advertise the annual percentage yield, or APY. \({ }^{7}\) In mathematical terms, this is the yearly percentage growth rate for the exponential function that models the account balance. In this exercise and the next, we study the relationship between the APR and the APY. We assume that the APR is \(10 \%\), or \(0.1\) as a decimal. To determine the APY when we know the APR, we need to know how often interest is compounded. For example, suppose for the moment that interest is compounded twice a year. Then to say that the APR is \(10 \%\) means that in half a year, the balance grows by \(\frac{10}{2} \%\) (or \(5 \%\) ). In other words, the \(\frac{1}{2}\) year percentage growth rate is \(\frac{0.1}{2}\) (as a decimal). Thus the \(\frac{1}{2}\)-year growth factor is \(1+\frac{0.1}{2}\). To find the yearly growth factor, we need to perform a unit conversion: One year is 2 half-year periods, so the yearly growth factor is \(\left(1+\frac{0.1}{2}\right)^{2}\), or \(1.1025\). a. What is the yearly growth factor if interest is compounded four times a year? b. Assume that interest is compounded \(n\) times each year. Explain why the formula for the yearly growth factor is $$ \left(1+\frac{0.1}{n}\right)^{n} $$ c. What is the yearly growth factor if interest is compounded daily? Give your answer to four decimal places.

Cell phones: The following table shows the number, in millions, of cell phone subscribers in the United States at the end of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Subscribers (millions) } \\ \hline 2001 & 128.4 \\ \hline 2002 & 140.8 \\ \hline 2003 & 158.7 \\ \hline 2004 & 182.1 \\ \hline 2005 & 207.9 \\ \hline \end{array} $$ a. Plot the natural logarithm of the data points. Does this plot make it look reasonable to approximate the original data with an exponential function? b. Find the regression line for the natural logarithm of the data and add its graph to the plot in part a. c. Construct an exponential model for the original subscribership data using the logarithm as a link.

Radioactive decay: A scientist is studying the amount of a radioactive substance present over a period of time. A plot of the logarithm of the amount shows a linear pattern. What type of function should the scientist use to model the original data?

Atmospheric pressure: The table below gives a measurement of atmospheric pressure, in grams per square centimeter, at the given altitude, in kilometers.17 $$ \begin{array}{|c|c|} \hline \text { Altitude } & \text { Atmospheric pressure } \\ \hline 5 & 569 \\ \hline 10 & 313 \\ \hline 15 & 172 \\ \hline 20 & 95 \\ \hline 25 & 52 \\ \hline \end{array} $$ (For comparison, 1 kilometer is about 0.6 mile, and 1 gram per square centimeter is about 2 pounds per square foot.) a. Plot the data on atmospheric pressure. b. Make an exponential model for the data on atmospheric pressure. c. What is the atmospheric pressure at an altitude of 30 kilometers? d. Find the atmospheric pressure on Earth鈥檚 surface. This is termed standard atmospheric pressure. e. At what altitude is the atmospheric pressure equal to 25% of standard atmospheric pressure?
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