91Ó°ÊÓ

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 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.

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 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.