Problem 76 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, \$ 3000\( is invested in an account at interest rate \)r$, compounded continuously. Find the time required for the amount to (a) double and (b) triple. $$ r=0.12 $$

Step-by-Step Solution

Answer

The time it takes for the investment to double is approximately 5.775 years and to triple is approximately 9.487 years

Step 1: Set up the equations

We need to set up two separate equations for when the amount doubles and when it triples. Given that the principal amount \( P \) is $3000 and the rate \( r \) is 0.12, for doubling, the amount \( A \) will be $6000 and for tripling it will be $9000. The two equations will be: \n\nFor doubling: \( 6000 = 3000 e^{0.12t_1} \) \n\nFor tripling: \( 9000 = 3000 e^{0.12t_2} \)

Step 2: Solve the equations

These are exponential equations, so you'd have to first divide each side by 3000 then take the natural logarithm on both sides to solve for \( t \).\n\nFor doubling: \n\nDivide both sides by 3000: \(2 = e^{0.12t_1}\) \n\nTake the natural logarithm of both sides: \( \ln 2 = 0.12t_1 \)\n\nSolve for \( t_1 \): \( t_1 = \frac{\ln 2}{0.12} \)\n\nFor tripling: \n\nDivide both sides by 3000: \(3 = e^{0.12t_2}\)\n\nTake natural logarithm of both sides: \( \ln 3 = 0.12t_2 \)\n\nSolve for \( t_2 \): \( t_2 = \frac{\ln 3}{0.12} \)

Step 3: Evaluate

Using a calculator, you'd find that when rounded to three decimal places, \( \frac{\ln 2}{0.12} \) is about 5.775 and \( \frac{\ln 3}{0.12} \) is about 9.487

Key Concepts

Exponential Equations

An exponential equation is one where variables are found in an exponent. In our exercise, the equations are used to determine how long it takes for an investment to grow to a certain amount. Since the interest is compounded continuously, we use the formula: \[ A = Pe^{rt} \]where:

• \( A \): the amount of money accumulated after time \( t \)
• \( P \): the principal amount (initial investment)
• \( r \): the rate of interest
• \( t \): time in years
• \( e \): the base of the natural logarithm, approximately equal to 2.718

In this exercise, exponential equations allow us to model how the principal grows. First, we'll set up equations to calculate the doubling and tripling time of the investment. Solving these equations often involves taking the natural logarithm to isolate the variable \( t \). Through simple division and logarithmic functions, the abstract concept becomes a manageable calculation.

Natural Logarithm

A natural logarithm is the power to which the constant \( e \) (approximately 2.718) must be raised to obtain a number. When working with exponential growth and interest formulas, we use the natural logarithm, denoted by \( \ln \). This function is crucial for dealing with exponential equations, as it "undoes" the exponential, helping us solve for the time variable \( t \).Here's a breakdown of how to use it:

• To solve the equation \( e^{0.12t} = 2 \), you would apply the natural logarithm to both sides, resulting in \( \ln(e^{0.12t}) = \ln(2) \).
• The property \( \ln(e^x) = x \) implies that \( \ln(e^{0.12t}) = 0.12t \).
• Thus, \( 0.12t = \ln(2) \), allowing us to isolate \( t \) by dividing by 0.12: \( t = \frac{\ln(2)}{0.12} \).

This process helps understand how long it will take for investments to grow under continuous compounding. The natural logarithm bridges exponential growth equations with tangible time solutions.

Interest Rate Calculations

Calculating interest with continuous compounding involves understanding the growth nature of money when interest is compounded at every possible moment. The continuous compounding formula \( A = Pe^{rt} \) brings in the mathematical constant \( e \) as a crucial part of these calculations.Compared to simple interest rates:

• Continuous compounding calculates more frequently, leading to slightly higher returns.
• The effect of the interest rate \( r \) is exponential rather than linear.
• Even small differences in \( r \) or \( t \) have a more noticeable impact on the future value.

By using the exponential function \( e^{rt} \), we find the amount \( A \) grows exponentially over time. As seen in the exercise, solving for \( t \) with specific target amounts (doubling or tripling) showcases the dynamic nature of continuous interest. This method highlights not just how much money grows but how interdependent the parameters of principal, rate, and time are in financial calculations. Understanding these relationships is key to evaluating investment opportunities effectively.