Problem 9 - Precalculus: Functions and Graphs

Question

An investment of \(P\) dollars increased to \(A\) dollars in \(t\) years. If interest was compounded continuously, find the interest rate. $$A=13,464, \quad P=1000, \quad t=20$$

Step-by-Step Solution

Answer

The interest rate is approximately 12.995%.

Step 1: Understand the Formula for Continuous Compounding

The formula for continuous compounding is given as \( A = P e^{rt} \), where \( A \) is the amount of money after time \( t \), \( P \) is the principal amount invested, \( r \) is the annual interest rate, and \( t \) is the time in years.

Step 2: Substitute the Known Values into the Formula

Substitute \( A = 13,464 \), \( P = 1000 \), and \( t = 20 \) into the continuous compounding formula. This gives us the equation: \[ 13,464 = 1000 imes e^{20r}. \]

Step 3: Solve for the Exponent

Divide both sides of the equation by 1000 to solve for the exponent, \( e^{20r} \): \[ 13.464 = e^{20r}. \]

Step 4: Apply Natural Logarithm to Both Sides

Take the natural logarithm of both sides to isolate \( 20r \): \[ \ln(13.464) = 20r. \]

Step 5: Calculate \( r \)

Calculate \( r \) by dividing both sides by 20: \[ r = \frac{\ln(13.464)}{20}. \] Use a calculator to find the natural logarithm and perform the division to find \( r \).

Step 6: Compute the Final Answer

Approximating \( \ln(13.464) \approx 2.599 \), divide by 20 to find:\[ r \approx \frac{2.599}{20} \approx 0.12995. \] Therefore, the interest rate \( r \) is approximately \( 0.12995 \), or 12.995%.

Key Concepts

Exponential Growth

Exponential growth is a way to describe how things increase rapidly over time. It happens when the growth rate of a mathematical function is proportional to the function's current value. This kind of growth is common in nature and finance.
In the context of continuous compounding for investments, the amount of money grows exponentially as it earns interest on both the initial principal and the accumulated interest over time. The formula used is \( A = P e^{rt} \), where:

• \(A\) is the amount after time \(t\)
• \(P\) is the initial principal
• \(r\) is the annual interest rate
• \(t\) is the time in years

This exponential formula shows that even a small interest rate can lead to significant growth if the investment period \(t\) is long enough. Exponential growth, therefore, powerfully demonstrates the benefit of time in investment returns.

Natural Logarithms

A natural logarithm is a logarithm with the base \(e\), an irrational constant approximately equal to \(2.71828\). It is denoted as \(\ln\).
Natural logarithms are the inverse of the exponential function, meaning they can "undo" what exponentials do.
For example, if you have \(e^x = y\), then \(\ln(y) = x\).
This property makes natural logarithms particularly useful for solving problems involving exponential growth, like those in continuous compounding.
In our continuous compounding formula \( 13.464 = e^{20r} \), we used natural logarithms to solve for the interest rate \(r\).
By applying the natural logarithm to both sides, \( \ln(13.464) = 20r \), we were able to isolate \(r\) and solve for it.
Understanding how to use natural logarithms is key when working with exponential equations such as those used for interest calculations.

Interest Rate Calculation

The interest rate in continuous compounding calculations reflects how quickly the investment grows.
To find this rate, rearrange the continuous compounding formula \( A = P e^{rt} \) to solve for \(r\).
First, simplify the equation: divide both sides by \(P\) to get \( e^{rt} = \frac{A}{P} \).
In our exercise, with \(A = 13,464\) and \(P = 1000\), this simplifies to \( e^{20r} = 13.464 \).
The next step is to use the natural logarithm to solve for \(r\). Taking the natural logarithm on both sides yields \( \ln(13.464) = 20r \).
By isolating \(r\), you can determine the interest rate that would allow the principal \(P\) to grow to \(A\) over the span of \(t\) years.
From the final step of the calculation, \( r = \frac{\ln(13.464)}{20} \), we arrive at an interest rate of approximately 12.995%.
Interest rate calculation in continuous compounding helps investors understand the rate at which their investment grows exponentially over time.