Problem 105 - College Algebra with Modeling and Visualization

Question

The data in the table satisfy the equation \(y=\frac{k}{x^{n}},\) where \(n\) is a positive integer. Determine \(k\) and \(n\) $$ \begin{array}{ccccc} x & 2 & 3 & 4 & 5 \\ \hline y & 1.5 & 1 & 0.75 & 0.6 \end{array} $$

Step-by-Step Solution

Answer

The values are \(k = 3\) and \(n = 1\).

Step 1: Recognize the Relationship

The table shows the relationship between \(x\) and \(y\) that satisfies the equation \(y = \frac{k}{x^n}\), where \(k\) is a constant and \(n\) is a positive integer.

Step 2: Calculate \(n\) using Two Data Points

Choose two pairs of \((x, y)\) values to determine \(n\). Using \((x_1, y_1) = (2, 1.5)\) and \((x_2, y_2) = (3, 1)\), apply the equation:\[ 1.5 = \frac{k}{2^n} \quad \text{and} \quad 1 = \frac{k}{3^n} \]Divide the equations:\[ \frac{1.5}{1} = \frac{3^n}{2^n} \]Simplify to:\[ 1.5 = \left(\frac{3}{2}\right)^n \]To find \(n\), equate:\[ n \cdot \log\left(\frac{3}{2}\right) = \log(1.5) \]

Step 3: Solve for \(n\)

With the log equation established:\[ n = \frac{\log(1.5)}{\log\left(\frac{3}{2}\right)} \]Using a calculator, find:\[ \log(1.5) \approx 0.1761, \quad \log\left(\frac{3}{2}\right) \approx 0.1761 \]So:\[ n \approx \frac{0.1761}{0.1761} \approx 1 \]Therefore, we conclude \(n = 1\).

Step 4: Determine \(k\) using One Data Point

Substitute \(n = 1\) back in the original equation using any point, say \(x = 2, y = 1.5\):\[ 1.5 = \frac{k}{2^1} \]Thus, \(k = 1.5 \times 2 = 3\).

Step 5: Verify the Solution with Other Points

Use the values \(k = 3\) and \(n = 1\) in the equation \(y = \frac{3}{x}\) for the other given pairs:For \(x = 3\), \(y = \frac{3}{3} = 1\); matches.For \(x = 4\), \(y = \frac{3}{4} = 0.75\); matches.For \(x = 5\), \(y = \frac{3}{5} = 0.6\); matches.The solution satisfies all data points.

Key Concepts

Constant of Variation

In inverse variation, the constant of variation is a fixed value that relates the two variables involved. For the equation \(y = \frac{k}{x^n}\), \(k\) represents the constant of variation. To find the constant of variation, we first determine the exponent \(n\) from the relationship between the given data points. Once \(n\) is known, use any of the

• Data points such as \((x, y)\) to substitute back into the equation \(y = \frac{k}{x^n}\)
• Solve for \(k\) by isolating it on one side of the equation

With the exercise provided, we determined that \(n = 1\). Coming back to find \(k\), use any pair like \((2, 1.5)\):\[ 1.5 = \frac{k}{2^1} \]Which gives us:\[ k = 1.5 \times 2 = 3 \]Hence, the constant of variation \(k\) is 3. This value remains the same for all other values \(x\) in the inverse variation relationship described by the equation.

Logarithmic Calculation

Logarithmic calculation is an essential tool in solving equations involving inverse variation. It helps isolate the variables and constants by transforming multiplicative relationships into additive ones. When solving for \(n\) in the equation \( y = \frac{k}{x^n} \), we can equate the ratio of product values using logarithms to simplify the calculation:- Start by dividing one equation by another from chosen data points and simplify the relationship to a power equation, \(1.5 = \left(\frac{3}{2}\right)^n\), for example.- To solve for \(n\), employ logarithms:\[ n \cdot \log\left(\frac{3}{2}\right) = \log(1.5)\]- Derive \(n\) by dividing:\[ n = \frac{\log(1.5)}{\log\left(\frac{3}{2}\right)}\]The use of logarithms linearizes the equation, making it easier to solve for unknowns. After using a calculator, if needed, you arrive at \(n = 1\), demonstrating an effective application of logarithmic calculation.

Data Point Analysis

Data point analysis involves examining specific values of variables given in a dataset to identify relationships or patterns. In inverse variation contexts, it is crucial for determining variables and verifying solutions. Analysis steps include:

• Select any two data points for comparison
• Use chosen points to simplify and solve for unknown parameters, such as \(n\)
• Verify results across all data points to ensure correctness

For instance, using the points \((x, y) = (2, 1.5)\) and \((3, 1)\), we determined \(n = 1\). With this result, substituting back into the equation verifies the constant and the relationship for other data values:- For \(x = 4\), the computed \(y = \frac{3}{4} = 0.75\), matching given data- For \(x = 5\), \(y = \frac{3}{5} = 0.6\), again matches given dataThis validation ensures the inverse variation equation correctly describes the dataset, reaffirming the accuracy of determining \(k\) and \(n\). Data point analysis is a systematic approach to ensure comprehensive understanding and verification of solutions.