Problem 38 - Intermediate Algebra

Question

A piece of machinery valued at $$\$ 30,000$$ depreciates at a rate of \(10 \%\) yearly. How long will it take until it has a value of $$\$ 15,000$$ ?

Step-by-Step Solution

Answer

The machinery will take approximately 7 years to depreciate to $15,000.

Step 1: Identify the Depreciation Formula

The value of an asset depreciating at a constant rate can be calculated with the formula: \( V = P(1 - r)^t \), where \( V \) is the future value, \( P \) is the initial value, \( r \) is the depreciation rate, and \( t \) is the time in years we want to find.

Step 2: Assign Known Values

We know that \( V = 15000 \), \( P = 30000 \), and \( r = 0.10 \). Our goal is to find \( t \) when the machine value becomes \$15,000.

Step 3: Set Up the Equation

Plug in the known values into the depreciation formula: \( 15000 = 30000(1 - 0.10)^t \). This simplifies to \( 15000 = 30000(0.90)^t \).

Step 4: Solve for \( t \)

First, divide both sides by 30000 to isolate the term with \( t \): \( \frac{15000}{30000} = (0.90)^t \). This simplifies to \( 0.5 = (0.90)^t \).

Step 5: Use Logarithms to Solve for \( t \)

Take the natural logarithm of both sides to solve for \( t \): \( \ln(0.5) = t \cdot \ln(0.90) \). Now, solve for \( t \) by dividing both sides by \( \ln(0.90) \): \( t = \frac{\ln(0.5)}{\ln(0.90)} \).

Step 6: Calculate \( t \)

Use a calculator to find \( \ln(0.5) \approx -0.6931 \) and \( \ln(0.90) \approx -0.1054 \). Substituting these values gives \( t \approx \frac{-0.6931}{-0.1054} \approx 6.58 \). Round \( t \) to the nearest whole number to find the years: \( t \approx 7 \) years.

Key Concepts

Depreciation Formula

Depreciation is a method used to allocate the cost of a tangible asset over its useful life. In simple terms, it's a way to spread out the expense of an asset over the years you use it. To calculate the depreciation of an asset like machinery, you can use the Depreciation Formula, which helps determine how much value an asset loses each year.

The formula is given by: \( V = P(1 - r)^t \), where:

• \( V \) is the future value of the asset, or its worth after a certain number of years.
• The initial value of the asset is represented by \( P \), which is its original cost.
• \( r \) is the depreciation rate, expressed as a decimal.
• The time in years, \( t \), is the unknown we're looking to calculate.

This formula is essential for businesses and individuals who need to track the decreasing value of their investments and decide when it's time to replace them.

Exponential Decay

Exponential decay refers to the process where a quantity decreases at a rate proportional to its current value, which we often see in contexts like radioactivity, cooling, and, in our example, asset depreciation. In depreciation, the exponential decay model is used because the asset loses value by a certain percentage each year, not a fixed amount.

This means each year, the depreciation amount is calculated based on the remaining value of the asset rather than the original value. For our exercise's machinery, the value shrinks continuously at a fixed rate of 10% every year.

Using exponential decay with depreciation helps provide a more accurate picture of an asset's loss in value over time, reflecting how depreciation impacts the remaining value and changes annually.

Logarithms

Logarithms are mathematical operations that help us work with exponential equations easily by transforming them into linear equations. They are particularly useful when dealing with exponential decay, like in depreciation calculations.

In our exercise, we need to find how many years (\( t \)) it will take for the machinery's value to reach a specific amount. We ended up with an equation \( 0.5 = (0.90)^t \). By taking the natural logarithm of both sides, we convert this equation to a format that allows us to solve for \( t \).

The logarithmic process goes like this:

• Take the natural log of both sides: \( \ln(0.5) = t \cdot \ln(0.90) \).
• This transforms our problem from an exponential one into a multiplication one.
• Finally, isolate \( t \) by dividing: \( t = \frac{\ln(0.5)}{\ln(0.90)} \).

This step is crucial because it gives us a straightforward path to find the number of years needed for the value to depreciate to the desired level.

Algebraic Equations

Algebra is an essential branch of mathematics that deals with symbols and the rules for manipulating those symbols. In this exercise, we need to solve an algebraic equation to find how long it will take for the asset's value to halve.

Initially, we set up the depreciation equation using known variables, resulting in:\( 15000 = 30000(0.90)^t \).

This algebraic equation needed adjustments to isolate \( t \), the variable we want to find:

• Start by dividing both sides by 30000: \( \frac{15000}{30000} = (0.90)^t \), which simplifies to \( 0.5 = (0.90)^t \).
• We used logarithms to solve for \( t \) since it's an exponent in this equation.

By addressing each mathematical step methodically, algebra helps decipher complex scenarios like depreciation, making it more manageable and less intimidating for students.