Problem 26 - Basic Engineering Mathematics

Question

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 \(£ 300\).

Step-by-Step Solution

Answer

(a) £215.89; (b) 14 years.

Step 1: 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, \( n = 1 \).

Step 2: 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.

Step 3: 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:\[ 100 \left(1 + \frac{0.08}{1}\right)^t > 300 \]\[ (1.08)^t > 3 \]

Step 4: 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.

Key Concepts

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.

In many cases, such as the one in our exercise, the interest is compounded annually. This simplifies our formula slightly since \( n = 1 \). The compound interest formula is so powerful because it accounts not only for interest on the initial principal but also for interest on the accumulated interest from previous periods. This is what distinguishes it from simple interest.

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 \( £215.89 \).

This outcome means your initial \(£100\) investment would grow to approximately \(£215.89\) over 10 years. It illustrates the power of compounding, where interest is effectively earning its own interest over time.

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:

• 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 \( t \): \[ t > \frac{\log(3)}{\log(1.08)} \approx 14.28 \]

Hence, it takes about 14 years to exceed \(£300\). This part of financial mathematics illustrates how logarithms can be used to solve exponential equations, providing insights into time frames for investment objectives.