Problem 63 - Algebra and Trigonometry

Question

Depreciation A construction company purchases a bulldozer for \(\$ 160,000\) . Each year the value of the bulldozer depreciates by 20\(\%\) of its value in the preceding year. Let \(V_{n}\) be the value of the bulldozer in the \(n\) th year. (Let \(n=1\) be the year the bulldozer is purchased.) (a) Find a formula for \(V_{n}\) . (b) In what year will the value of the bulldozer be less than \(\$ 100,000 ?\)

Step-by-Step Solution

Answer

(a) \( V_{n} = 160,000 \times (0.8)^{n-1} \); (b) Value below $100,000 in year 6.

Step 1: Understanding Depreciation

The bulldozer depreciates by 20% each year. This means each year, its value is reduced to 80% of the previous year's value. For year 1, the value is the purchase price, which is $160,000.

Step 2: Establish Recurrence Relation for Depreciation

To establish how the bulldozer's value changes each year, note that each year's value is given by multiplying the previous year's value by 0.8 (since 100% - 20% = 80%). This gives us the recurrence relation: \( V_{n} = 0.8 \times V_{n-1} \).

Step 3: Derive the Formula for Value Over Time

Starting with the initial value, \( V_{1} = 160,000 \), we can express the value in any year \( n \) as \( V_{n} = 160,000 \times (0.8)^{n-1} \). This formula uses the fact that each year the bulldozer retains 80% of its previous year's value.

Step 4: Find When Value Drops Below $100,000

We need to find the smallest \( n \) such that \( V_{n} < 100,000 \). Start by setting up the inequality with our formula: \( 160,000 \times (0.8)^{n-1} < 100,000 \).

Step 5: Solve the Inequality for n

First, simplify the inequality: \( (0.8)^{n-1} < \frac{100,000}{160,000} = \frac{5}{8} \). Take the log of both sides to solve for \( n \): \( \log((0.8)^{n-1}) < \log(\frac{5}{8}) \). Using logarithmic properties, this becomes \( (n-1) \cdot \log(0.8) < \log(\frac{5}{8}) \). Divide both sides by \( \log(0.8) \) (noting that \( \log(0.8) \) is negative, which reverses the inequality): \( n-1 > \frac{\log(\frac{5}{8})}{\log(0.8)} \).

Step 6: Calculate n

Compute the right side using a calculator: \( \log(\frac{5}{8}) \approx -0.09691 \) and \( \log(0.8) \approx -0.09691 \). Thus, \( n-1 > 4.3219 \), giving \( n > 5.3219 \). Since \( n \) must be a whole number, the smallest integer \( n \) is 6.

Key Concepts

Recurrence Relations

In algebra, recurrence relations help us describe situations where the current value depends on prior values. For the bulldozer's depreciation, each year the value decreases to 80% of the previous year's value. This creates a recurrence relationship using a formula:

• \( V_{n} = 0.8 \times V_{n-1} \)

The concept is straightforward: use the known percentage decrease (or the factor, 0.8 in this case) to establish the next term from the previous term. Start with the initial value of the bulldozer, which is \( V_{1} = 160,000 \). Then use the relation to find successive values by repeatedly applying the formula. Recurrence relations like this are common in finance-related arithmetic sequence problems, where objects lose value steadily over time. They simplify real-world processes by showing value changes step-by-step in a predictable manner, which is crucial when it comes to calculations like depreciation.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. Unlike simple subtraction, exponential decay involves multiplying the current value by a constant factor less than 1. The bulldozer depreciate at an exponential rate because each year, its value is multiplied by 0.8.

• The formula for the value of the bulldozer over time becomes: \( V_{n} = 160,000 \times (0.8)^{n-1} \)

In this formula:

• \( 160,000 \) represents the initial value of the bulldozer
• \( 0.8 \) is the decay factor, as the bulldozer maintains only 80% of its value each year
• \( n-1 \) accounts for the number of complete years since the purchase

This type of mathematical description is pivotal not only in finance but in various natural phenomena where exponential decrements appear, such as radioactive decay or population decline. The essence of exponential decay in algebraic form helps to predict future values under consistent percentage reductions.

Inequality Solving

To find when a depreciating asset falls below a certain value, we solve inequalities. For the bulldozer, we need the year when its value drops below $100,000. The equation is set up as:

• \( 160,000 \times (0.8)^{n-1} < 100,000 \)

Break it down by simplifying:

• Firstly, \( (0.8)^{n-1} < \frac{5}{8} \), after dividing both sides by 160,000

Solving this inequality involves logarithms, transforming the base-0.8 exponential expression into something we can handle algebraically. Remember:

• When dividing or multiplying by negative in inequalities, the sign flips

The process highlights key algebraic manipulation techniques used to find solutions satisfying the condition of a certain range or limit, making inequality solving indispensable in finding when depreciation targets are met.

Logarithms

Logarithms convert multiplicative relationships into additive ones, making them extremely useful in exponential situations like depreciation. Solving the bulldozer's inequality \( (0.8)^{n-1} < \frac{5}{8} \) requires taking the logarithm of both sides:

• \( \log((0.8)^{n-1}) < \log(\frac{5}{8}) \)

This transformation simplifies it to:

• \( (n-1) \cdot \log(0.8) < \log(\frac{5}{8}) \)

The properties of logarithms allow us to move exponents out front, aiding in solving for \( n \). With calculators, the usage of logarithms becomes practical as they convert complex exponential equations into linear forms:

• Calculate \( \log(0.8) \approx -0.09691 \) and \( \log(\frac{5}{8}) \approx -0.09691 \)
• Solve \( n-1 > \frac{\log(\frac{5}{8})}{\log(0.8)} \)

Understanding how logarithms change the form of the equation offers a doorway into more complex algebraic analysis, crucial for solving depreciation scenarios involving exponential decay.