Chapter 29: Problem 26
Problem 26. If \(£ 100\) is invested at compound interest of \(8 \%\) per annum, determine (a) the value after 10 years,(b) the time, correct to the nearest year, it takes to reach more than \(£ 300\).
Short Answer
Expert verified
(a) £215.89; (b) 14 years.
Step by step solution
01
Understand Compound Interest Formula
The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount after time \( t \),- \( P \) is the principal amount (initial investment),- \( r \) is the annual interest rate (as a decimal),- \( n \) is the number of times the interest is compounded per year,- \( t \) is the time in years.For this problem, since interest is compounded annually, \( n = 1 \).
02
Calculate the Value After 10 Years
Given: \( P = 100 \), \( r = 0.08 \), \( n = 1 \), \( t = 10 \).Substitute these values into the formula:\[ A = 100 \left(1 + \frac{0.08}{1}\right)^{1 \times 10} \] \[ A = 100 (1.08)^{10} \] \[ A \approx 215.89 \] So, the value after 10 years is approximately £215.89.
03
Setup the Equation for Time to Reach Over £300
We need \( A > 300 \), given \( P = 100 \), \( r = 0.08 \), and \( n = 1 \). We want to find the time \( t \) when:\[ 100 \left(1 + \frac{0.08}{1}\right)^t > 300 \]\[ (1.08)^t > 3 \]
04
Solve for Time \( t \) Using Logarithms
Take the logarithm of both sides:\[ \log((1.08)^t) > \log(3) \]Using the property \( \log(a^b) = b \log(a) \):\[ t \log(1.08) > \log(3) \]\[ t > \frac{\log(3)}{\log(1.08)} \]\[ t \approx \frac{0.4771}{0.03342} \approx 14.28 \]Therefore, it takes approximately 14 years to reach more than £300.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Compound Interest Formula
The compound interest formula is fundamental in financial mathematics, providing a way to calculate the future value of an investment based on interest that is compounded over time. The formula is represented as: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]where:
- \( A \) is the amount after time \( t \), meaning the total value of the investment at the end of the period.
- \( P \) is the principal amount, or the initial amount invested.
- \( r \) stands for the annual interest rate, expressed as a decimal.
- \( n \) is the number of times the interest is compounded per year.
- \( t \) represents the time in years.
Investment Growth Calculation
Calculating the growth of an investment using the compound interest formula involves a few straightforward steps. Suppose you have an initial investment of \(£100\), an annual interest rate of \(8\%\), and a time period of \(10\) years. To find out how much the investment will be worth at the end of these 10 years, you would:
- Convert the annual interest rate from percentage to decimal by dividing by 100, giving \( r = 0.08 \).
- Since the interest compounds annually, set \( n = 1 \).
- Substitute these values into the formula: \[ A = 100 \left(1 + \frac{0.08}{1}\right)^{1 \times 10} \]
- Simplify to find \( A = 100 (1.08)^{10} \), which calculates to approximately \( £215.89 \).
Financial Mathematics Concepts
Financial mathematics encompasses various concepts, but compound interest is one of the most critical because it influences everything from personal savings to large-scale investment strategies. Understanding how long it takes for an investment to grow to a desired amount involves solving equations using logarithms.
For instance, if an investor wants to know how long it will take for their \(£100\) investment to exceed \(£300\) at an \(8\%\) compound interest rate, the process is as follows:
For instance, if an investor wants to know how long it will take for their \(£100\) investment to exceed \(£300\) at an \(8\%\) compound interest rate, the process is as follows:
- Start with the inequality determination: \[ 100 \left(1 + \frac{0.08}{1}\right)^t > 300 \]
- Simplify to: \[ (1.08)^t > 3 \]
- Take the logarithm of both sides to facilitate solving for \( t \): \[ \log((1.08)^t) > \log(3) \]
- Using logarithmic properties, rewrite as: \[ t \cdot \log(1.08) > \log(3) \]
- Finally, solve for \( t \): \[ t > \frac{\log(3)}{\log(1.08)} \approx 14.28 \]