Problem 26 - Elementary and Intermediate Algebra

Question

In Exercises 25 and \(26,\) determine the time necessary for \(\$ 1000\) to double if it is invested at interest rate \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. \((m+4)(m+4)\)

Step-by-Step Solution

Answer

Use the formula \( t = \frac{\ln(2)}{n \cdot \ln(1 + \frac{r}{n})} \) for different \( n \). For continuous compounding, \( t = \frac{\ln(2)}{r} \).

Step 1: Understanding the problem

We need to determine the time it takes for an investment to double using compound interest for different compounding frequencies: annually, monthly, daily, and continuously. The principal amount is $1000, so we need to find when it becomes $2000 for each case.

Step 2: Formula for Compound Interest

The formula for compound interest when compounded periodically is given by: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]where \( A \) is the amount of money accumulated after n years, including interest. \( P \) is the principal amount, \( r \) is the annual interest rate (as a decimal), \( n \) is the number of times interest is compounded per year, and \( t \) is the time in years.

Step 3: Setting up the equation for doubling

To find the time for the money to double, set \( A = 2000 \) (since \( P = 1000\)), and substitute into our formula:\[ 2000 = 1000 \left(1 + \frac{r}{n}\right)^{nt} \]Simplifying, we get:\[ 2 = \left(1 + \frac{r}{n}\right)^{nt} \]

Step 4: (a) Compounding Annually

When compounding annually, \( n = 1 \). Substitute \( n = 1 \) into the equation:\[ 2 = \left(1 + r\right)^t \]To solve for \( t \), take the natural logarithm of both sides:\[ \ln(2) = t \cdot \ln(1 + r) \]So, \( t = \frac{\ln(2)}{\ln(1 + r)} \). This provides the time to double when compounded annually.

Step 5: (b) Compounding Monthly

For monthly compounding, \( n = 12 \). Substitute \( n = 12 \) into the equation:\[ 2 = \left(1 + \frac{r}{12}\right)^{12t} \]Take the logarithm:\[ \ln(2) = 12t \cdot \ln\left(1 + \frac{r}{12}\right) \]Solve for \( t \):\[ t = \frac{\ln(2)}{12 \cdot \ln\left(1 + \frac{r}{12}\right)} \]

Step 6: (c) Compounding Daily

For daily compounding, typically \( n = 365 \). Substitute \( n = 365 \) into the equation:\[ 2 = \left(1 + \frac{r}{365}\right)^{365t} \]Take the logarithm:\[ \ln(2) = 365t \cdot \ln\left(1 + \frac{r}{365}\right) \]Solve for \( t \):\[ t = \frac{\ln(2)}{365 \cdot \ln\left(1 + \frac{r}{365}\right)} \]

Step 7: (d) Continuous Compounding

For continuous compounding, we use the formula:\[ A = Pe^{rt} \]Substitute in for \( A \) and \( P \):\[ 2000 = 1000e^{rt} \]\[ 2 = e^{rt} \]Taking the natural logarithm:\[ \ln(2) = rt \]So, \( t = \frac{\ln(2)}{r} \). This provides the time to double with continuous compounding.

Key Concepts

Annual Compounding

Annual compounding refers to the process where interest is calculated and added to the principal balance once a year. It's one of the simplest and most straightforward methods of compounding. Let's break it down:

• The principal amount, in this case, is the original sum of money invested, which is $1000.
• The annual interest rate is applied once per year.
• To calculate the time needed for the investment to double, the formula is: \[ 2 = (1 + r)^t \]Here, \( r \) is the annual interest rate expressed as a decimal, and \( t \) represents time in years.

To find \( t \), simply solve: \[ t = \frac{\ln(2)}{\ln(1 + r)} \]The logarithmic approach here is used to determine how many time periods are required to reach the desired amount, making it a useful formula for strategic financial planning.

Monthly Compounding

Monthly compounding increases the frequency at which interest is calculated and added to the account balance. As the name suggests, this occurs every month:

• The principal remains the initial $1000.
• The annual interest rate is divided by 12, corresponding to monthly intervals.
• Using the compound interest formula, we aim to find the time "t" when the investment doubles: \[ 2 = \left( 1 + \frac{r}{12} \right)^{12t} \]

By taking the logarithm of both sides, you simplify and solve: \[ t = \frac{\ln(2)}{12 \cdot \ln\left(1 + \frac{r}{12}\right)} \]Monthly compounding typically yields a higher total than annual compounding, due to how regularly interest is added. This option is often preferable for investments anticipating significant compounding over time.

Daily Compounding

With daily compounding, interest is calculated and added every day. This means 365 compounding periods are considered each year:

• The starting $1000 remains our principal.
• The interest rate gets divided by 365, attributing one part to each day of the year.
• We check when the investment value doubles by setting up: \[ 2 = \left( 1 + \frac{r}{365} \right)^{365t} \]

Use the logarithmic approach to solve for time: \[ t = \frac{\ln(2)}{365 \cdot \ln\left(1 + \frac{r}{365}\right)} \]Daily compounding results in a slightly larger accumulation than monthly compounding over the same period, illustrating how frequent compounding can enhance financial growth.

Continuous Compounding

Continuous compounding applies the concept of compounding interest at every possible instant. It's a mathematical idealization using the constant \( e \), representing exponential growth:

• Here, the principal of $1000 is employed.
• The compounding formula generalizes to: \[ A = Pe^{rt} \]
• We solve for when the investment doubles: \[ 2 = e^{rt} \]

Find time by: \[ t = \frac{\ln(2)}{r} \]This formula shows the rapid effect of continuous compounding, often yielding the highest returns over the same period when compared to periodic methods. Continuous compounding is ideal for understanding potential growth in an ideal environment where compounding is limitless.