Problem 77 - Precalculus : Building Concepts and Connections

Question

The value of a 2006 S-type Jaguar is given by the function $$v(t)=43,173(0.8)^{t}$$ where \(t\) is the number of years since its purchase and \(v(t)\) is its value in dollars. (Source: Kelley Blue Book) (a) What was the Jaguar's initial purchase price? (b) What percentage of its value does the Jaguar S-type lose each year? (c) How many years will it take for the Jaguar S-type to reach a value of \(\$ 22,227 ?\)

Step-by-Step Solution

Answer

The initial cost of the car was $43,173. The car depreciates by 20% annually. The required time for the car to drop to a value of $22,227 is solved as \(\frac{\ln(22,227 / 43,173)}{\ln(0.8)}\) years. The exact value is dependent on the properties of natural logarithms and should be calculated using a tool capable of the such.

Step 1: Finding the Initial Cost of Car

From the given formula \(v(t)=43,173(0.8)^{t}\), we see the initial price is the value when \(t = 0\) (as \(t\) corresponds to the number of years). The initial price of the Jaguar is then given by \(v(0) = 43,173(0.8)^{0} = 43,173\). We evaluate the exponent as \(0.8^{0}\) which is equal to 1.

Step 2: Finding the Annual Depreciation Rate

The depreciation rate can be found directly from the formula also. In this case, the factor of the exponent, 0.8, is the multiplicative rate of change each year. To find the annual depreciation in percent, subtract this rate from 1 and multiply by 100 to convert to a percentage. \(Percentage\: lost\: each\: year = (1 - 0.8) * 100\% = 20\%\). The car is depreciating by 20% each year.

Step 3: Finding the Depreciation Time

To find the number of years it will take for the Jaguar S-type to reach $22,227, we need to solve the equation \(v(t) = 22,227\) for \(t\). That is, \(43,173(0.8)^{t} = 22,227\). First, divide both sides by 43,173 to isolate \(0.8^{t}\). We then take the natural logarithm (ln) of both sides. Using the property that the natural logarithm and exponentiation are inverse functions, we can simplify the equation to find \(t\). So, \(t\) will be \(\frac{\ln(22,227 / 43,173)}{\ln(0.8)}\). Evaluating the right hand side will give the correct value of \(t\) in years.

Key Concepts

Depreciation

Depreciation, in the context of this exercise, refers to the reduction in the car's value over time. Specifically, over each year, the Jaguar S-type loses a portion of its value, a common phenomenon with cars, which typically depreciate due to wear and tear, and market conditions.
\[\]The formula given, \(v(t) = 43,173(0.8)^{t}\), demonstrates exponential decay, where each year (\(t\)) results in the car losing a consistent percentage of its value. The depreciation rate here is the result of the formula factor 0.8.

• Initial Value: When \(t=0\), the Jaguar was purchased at the full price of \$43,173, with no depreciation applied yet.
• Annual Depreciation: This is calculated by subtracting the exponential factor from 1, giving us \(1 - 0.8 = 0.2\), which corresponds to 20% per year.

Therefore, each year, the car retains 80% of its value from the previous year, leading to a 20% loss annually. This concept is crucial in financial planning, as understanding depreciation can help manage asset values better and make more informed economic decisions.

Mathematical Modeling

Mathematical modeling is a real-world application of mathematics that uses equations and formulas to represent and solve problems accurately. The function \(v(t) = 43,173(0.8)^{t}\) models the car's value depreciation over time using an exponential decay formula.
\[\]This type of model was employed because:

• Predictive Power: By inputting any given year into the equation, one can predict the car's worth effectively.
• Realism: The exponential decay adequately reflects how vehicles typically decrease sharply in value initially and then taper off.
• Communicability: Using such functions provides a straightforward way to communicate and evaluate real-world situations mathematically.

This model simplifies complex real-world factors into a digestible mathematical description, granting both clarity and accuracy for financial and economic analyses. With such models, individuals and businesses can forecast future value, aiding in investment and sale decisions.

Logarithmic Functions

To solve how many years it will take for the Jaguar's value to decrease to \$22,227, logarithmic functions play a key role. When you solve the equation \(43,173(0.8)^{t} = 22,227\), you make use of logarithms to isolate \(t\).
\[\]Here, the approach involves:

• Isolating the Exponent: Divide both sides by 43,173 to address \(0.8^{t}\).
• Using Logarithms: Employ natural logs: \(\ln(0.8^{t})\) = \(\ln(22,227 / 43,173)\). This simplifies the exponential form.
• Simplifying: With log properties, simplify to find \(t = \frac{\ln(22,227 / 43,173)}{\ln(0.8)}\).

By applying logarithms to exponential functions, you can solve for unknowns like time, transforming otherwise complex calculations into manageable steps. Logarithmic functions thus provide an essential tool for tackling exponential decay problems where direct calculation isn’t possible.