Problem 52
Question
COMPOUND INTEREST A sum of money is invested at a certain fixed interest rate, and the interest is compounded continuously. After 10 years, the money has doubled. How will the balance at the end of 20 years compare with the initial investment?
Step-by-Step Solution
Verified Answer
The balance at the end of 20 years will be 4 times the initial investment.
1Step 1 - Understand the Formula
For continuous compounding, use the formula: \[ A = P e^{rt} \] where \( A \) is the amount of money accumulated after time \( t \), \( P \) is the principal amount (initial investment), \( r \) is the interest rate, and \( t \) is the time in years.
2Step 2 - Establish the Relationship for 10 Years
Given that the money doubles in 10 years: \[ 2P = P e^{10r} \] Dividing both sides by \( P \) gives: \[ 2 = e^{10r} \]
3Step 3 - Solve for the Interest Rate, r
To find the interest rate \( r \), take the natural logarithm of both sides: \[ \ln(2) = 10r \] Then, solve for \( r \): \[ r = \frac{\ln(2)}{10} \]
4Step 4 - Calculate the Balance at 20 Years
Now use the formula again to find the balance at 20 years: \[ A = P e^{rt} \] Substitute \( r = \frac{\ln(2)}{10} \) and \( t = 20 \): \[ A = P e^{20 \cdot \frac{\ln(2)}{10}} = P e^{2 \ln(2)} = P (e^{\ln(2)})^2 = P (2)^2 = 4P \]
Key Concepts
continuous compoundinginvestment growthnatural logarithminterest rate calculation
continuous compounding
Continuous compounding is a powerful concept in finance that affects how investments grow. Instead of compounding interest at regular intervals like monthly or annually, continuous compounding calculates interest that is added constantly.
Imagine if your bank account added tiny amounts of interest every single moment. With continuous compounding, your money grows faster than with standard compounding methods.
The formula for continuous compounding is:
equations with actual characters there can be put inside latex equations with actual characters there can be put inside latex\( A = Pe^{rt} \).
Imagine if your bank account added tiny amounts of interest every single moment. With continuous compounding, your money grows faster than with standard compounding methods.
The formula for continuous compounding is:
equations with actual characters there can be put inside latex equations with actual characters there can be put inside latex\( A = Pe^{rt} \).
- A is the amount of money after time t
- P is the initial investment (principal)
- e is Euler's number, about 2.718
- r is the annual interest rate
- t is the time in years
investment growth
Understanding how investments grow over time is crucial. When interest is compounded continuously, investments can grow exponentially. This exercise shows that an initial amount of money can double in a specific period due to continuous compounding.
By using the formula for continuous compounding, we saw the money double after 10 years:
\( 2P = Pe^{10r} \).
After 20 years, applying the same principle, the initial investment grows to four times its original amount:
\( A = P e^{2 \text{ ln}(2)} = 4P \).
In simple terms:
By using the formula for continuous compounding, we saw the money double after 10 years:
\( 2P = Pe^{10r} \).
After 20 years, applying the same principle, the initial investment grows to four times its original amount:
\( A = P e^{2 \text{ ln}(2)} = 4P \).
In simple terms:
- Double your investment in 10 years
- Quadruple your investment in 20 years
natural logarithm
The natural logarithm, denoted as \( \text{ln} \), is essential in continuous compounding calculations. It is the inverse function of taking a number to the power of \( e \), meaning it helps to undo exponential growth.
When solving for the interest rate \( r \) in our exercise, we find:\( \text{ln}(2) = 10r \).
To get \( r \), divide both sides by 10:
\( r = \frac{\text{ln}(2)}{10} \).
The natural logarithm helps us make sense of exponential growth and is a key part of calculating continuous compound interest.
Using \( \text{ln} \) helps unravel the mysteries of growth rates, making financial calculations accurate and reliable.
When solving for the interest rate \( r \) in our exercise, we find:\( \text{ln}(2) = 10r \).
To get \( r \), divide both sides by 10:
\( r = \frac{\text{ln}(2)}{10} \).
The natural logarithm helps us make sense of exponential growth and is a key part of calculating continuous compound interest.
Using \( \text{ln} \) helps unravel the mysteries of growth rates, making financial calculations accurate and reliable.
interest rate calculation
Calculating the interest rate from continuous compounding involves understanding the relationship between growth factors and time. In the exercise, we used given information to find the rate:
\( 2 = e^{10r} \)
Taking the natural logarithm of both sides:\( \text{ln}(2) = 10r \).
Therefore:\( r = \frac{\text{ln}(2)}{10} \).
This rate tells us how fast our investment grows annually when compounded continuously.
Continuous compounding interest rates are valuable tools in finance, helping to quantify how time affects money growth. Calculating these rates accurately ensures better financial planning and decision-making.
\( 2 = e^{10r} \)
Taking the natural logarithm of both sides:\( \text{ln}(2) = 10r \).
Therefore:\( r = \frac{\text{ln}(2)}{10} \).
This rate tells us how fast our investment grows annually when compounded continuously.
Continuous compounding interest rates are valuable tools in finance, helping to quantify how time affects money growth. Calculating these rates accurately ensures better financial planning and decision-making.
Other exercises in this chapter
Problem 47
Find \(f(9)\) if \(f(x)=e^{k x}\) and \(f(3)=2\).
View solution Problem 51
RADIOACTIVE DECAY A radioactive substance decays exponentially. If 500 grams of the substance were present initially and 400 grams are prescnt 50 years later, h
View solution Problem 53
GROWTH OF BACTERIA The following data were compiled by a researcher during the first 10 minutes of an experiment designed to study the growth of bacteria: \begi
View solution Problem 54
RADIOACTIVE DECAY The following data were compiled by a researcher during an experiment designed to study the decay of a radioactive substance: \begin{tabular}{
View solution