Problem 56 - Introductory and Intermediate Algebra for College Students 4th

Question

The formula \(S=C(1+r)^{t}\) models inflation, where \(C=\) the value today, \(r=\) the anmual inflation rate, and \(S=\) the inflated value \(t\) years from now. Use this formula to solve. Round answers to the nearest dollar. If the inflation rate is \(3 \%,\) how much will a house now worth \(\$ 510,000\) be worth in 5 years?

Step-by-Step Solution

Answer

The house which is worth \(\$510,000\) now, will be worth approximately \(\$594,000\) in 5 years if the inflation rate continues to be \(3\%\).

Step 1 - Convert Percentage into Decimal

Convert the annual inflation rate from a percentage to a decimal by dividing by 100. So, \(r = \frac{3}{100} = 0.03\)

Step 2 - Insert into Formula

Insert the given values into the formula \(S = C(1+r)^{t}\), replacing \(C = \$510,000\), \(r = 0.03\), and \(t = 5\). This gives \(S = 510000 \times (1+0.03)^{5}\)

Step 3 - Calculate Inflated Value

First perform the operation in brackets \(1+0.03 = 1.03\). Then apply the power \(5\) to \(1.03\). Lastly, multiply the result with \$510,000 to get the inflated value. The usage of a calculator will make it easier to perform these operations.

Step 4 - Round to nearest dollar

After calculating the inflated value, round the result to the nearest dollar, because the context of the problem is about financial calculations. Any decimals less than \$0.50 are rounded down, while \$0.50 and above are rounded up

Key Concepts

Algebraic Problem Solving

Understanding how to approach algebraic problems is essential in many fields of study, including economics. When faced with a formula like S=C(1+r)^{t}, it's crucial to have a systematic approach. The first step is identifying the known variables. In our case, C represents the current value or cost, r is the annual inflation rate, and t is the time in years. To solve the problem effectively, break it down into manageable parts: convert percentages to decimals, plug in the values, and simplify the expression. Algebraic problem solving often requires these conversions and substitute operations to simplify complex problems into solvable equations.

For the given exercise, we replaced the known quantities into the inflation formula, considering the inflation rate as a decimal, which then made the exponential calculations straightforward. By using the systematic approach, students can not only solve this problem but also tackle similar algebraic equations in various real-world scenarios.

Exponential Functions

In our daily lives, exponential functions describe situations where something grows or decays at a rate proportional to its current value. The formula S=C(1+r)^{t} represents such a situation, with the amount of growth determined by the exponent t.

Understanding that the base of the exponential function, in this case, (1+r), remains constant while the exponent changes, is the key to grasping exponential growth. In our problem, we used the exponent 5 to reflect the 5-year period. By calculating the value of (1+0.03)^5, students learn the impact of exponential growth over time. This concept of exponential functions not only applies to economics and finance but also in sciences like biology and physics, making it a critical mathematical concept to comprehend.

Percentage to Decimal Conversion

Working with percentages is a common requirement in various fields, from finance to engineering. In many mathematical problems, we need to convert percentages to decimals to perform calculations. To do this, we divide the percentage value by 100. The reason for this conversion is that 'percent' means 'per hundred', so we are essentially finding the equivalent value for one, which is the basic unit for our decimal system.

In the context of our inflation rate problem, we converted the 3% inflation rate to 0.03 by dividing it by 100. It is a small but crucial step that impacts all subsequent calculations. Remembering that 1% equals 0.01 in decimal form is a simple tip that can help students with quick mental conversions. This concept is foundational for understanding interest rates, growth rates, and any scenario where percentage change occurs.