Problem 115 - Elementary and Intermediate Algebra

Question

The logarithmic function \(c(m)=500 \log (m+1)\) approximates the total number of cases handled one year by the staff at a county department of social services. The variable \(m\) represents the month of the year, where January is 1, February is 2, and so on. Find the total number of cases handled by the department that year by the end of June.

Step-by-Step Solution

Answer

Total cases by end of June are approximately 1851.

Step 1: Identify the Range of Months

Since we need to find the total number of cases handled by the end of June, we consider the months from January (\(m=1\)) to June (\(m=6\)).

Step 2: Calculate Total Cases for Each Month

We apply the given function \(c(m) = 500 \log(m+1)\) to each month from January to June and find the total number of cases for these months:- For January (\(m=1\)): \(c(1) = 500 \log(1+1) = 500 \log 2\)- For February (\(m=2\)): \(c(2) = 500 \log(2+1) = 500 \log 3\)- For March (\(m=3\)): \(c(3) = 500 \log(3+1) = 500 \log 4\)- For April (\(m=4\)): \(c(4) = 500 \log(4+1) = 500 \log 5\)- For May (\(m=5\)): \(c(5) = 500 \log(5+1) = 500 \log 6\)- For June (\(m=6\)): \(c(6) = 500 \log(6+1) = 500 \log 7\)

Step 3: Compute Each Logarithmic Value

Calculate the value for each logarithm using a calculator.- \(\log 2 \approx 0.3010\), so \(c(1) = 500 \times 0.3010 = 150.5\)- \(\log 3 \approx 0.4771\), so \(c(2) = 500 \times 0.4771 = 238.55\)- \(\log 4 \approx 0.6021\), so \(c(3) = 500 \times 0.6021 = 301.05\)- \(\log 5 \approx 0.6990\), so \(c(4) = 500 \times 0.6990 = 349.5\)- \(\log 6 \approx 0.7782\), so \(c(5) = 500 \times 0.7782 = 389.1\)- \(\log 7 \approx 0.8451\), so \(c(6) = 500 \times 0.8451 = 422.55\)

Step 4: Sum the Total Cases for the Six Months

Add up the values calculated for each of the six months:\[ c(1) + c(2) + c(3) + c(4) + c(5) + c(6) = 150.5 + 238.55 + 301.05 + 349.5 + 389.1 + 422.55 = 1851.25 \]

Step 5: Interpret the Result

The total number of cases handled by the department by the end of June is approximately 1851.

Key Concepts

Applications of Logarithms

In various fields, logarithms serve as a useful mathematical tool. They help in simplifying complex calculations, especially when dealing with multiplicative processes. In this exercise, the logarithmic function models the growth in the number of cases processed by a county's social services department over different months.
By using the function \(c(m) = 500 \log(m+1)\), each month from January to June is computed to estimate the caseload. This model:

• Captures the relationship between time (months) and workload (cases) efficiently.
• Demonstrates how the rate of increase in cases might slow down over time due to the nature of logarithmic growth.
• Provides a manageable way to forecast and anticipate resource needs throughout the year.

Such applications of logarithms offer useful insights into planning and evaluating services in both public administration and business operations.

Algebraic Problem Solving

Algebraic methods often involve substituting given values into functions or equations to solve problems. In this scenario, the exercise simplifies a real-world problem into an algebraic expression.
Here's how this was achieved:

• The range of months from January to June is linked to corresponding values of \(m=1\) to \(m=6\).
• Each month is inputted into the logarithmic function to determine the number of cases.
• Utilizing simple algebraic calculations, the monthly workload is meticulously estimated.

This approach shows how translating real-world data into algebraic terms enables precise solutions, enhancing problem-solving skills fundamental to mathematics and its applications.

Interpreting Mathematical Results

Understanding results from mathematical calculations provides valuable insights. The output of our logarithmic function gives specific values for the number of cases managed each month up to June.
Here's what we arrive at:

• The computation of logarithmic values like \(\log 2\), \(\log 3\), etc., translates to specific quantities which, when summed, provide a cumulative figure.
• The final result of approximately 1851 cases reflects the total effort involved in handling departmental duties for half the year.
• Interpretation of this total underlines the importance of such analysis in planning for the remaining months of the year, potentially informing adjustments in workforce allocation or resource procurement.

Recognizing and accurately interpreting these mathematical results equips decision-makers to analyze data outcomes effectively, reinforcing strategic developments in resource management.