Problem 81 - Algebra and Trigonometry

Question

Investment The time required to double the amount of an investment at an interest rate \(r\) compounded continuously is given by $$t=\frac{\ln 2}{r}$$ Find the time required to double an investment at \(6 \%, 7 \%,\) and 8\(\% .\)

Step-by-Step Solution

Answer

11.55 years for 6%, 9.90 years for 7%, 8.66 years for 8%.

Step 1: Understand the Formula

The formula given is for the time required to double an investment with continuous compounding interest: \[ t = \frac{\ln 2}{r} \] where \( t \) is the time in years and \( r \) is the interest rate expressed as a decimal.

Step 2: Convert Percentage to Decimal

Convert the interest rates from percentage to decimal for calculation: \(6\% = 0.06\), \(7\% = 0.07\), and \(8\% = 0.08\).

Step 3: Calculate for 6% Interest Rate

Use the formula with \( r = 0.06 \): \[ t = \frac{\ln 2}{0.06} \approx \frac{0.693}{0.06} \approx 11.55 \] years.

Step 4: Calculate for 7% Interest Rate

Use the formula with \( r = 0.07 \): \[ t = \frac{\ln 2}{0.07} \approx \frac{0.693}{0.07} \approx 9.90 \] years.

Step 5: Calculate for 8% Interest Rate

Use the formula with \( r = 0.08 \): \[ t = \frac{\ln 2}{0.08} \approx \frac{0.693}{0.08} \approx 8.66 \] years.

Key Concepts

Investment Doubling Time

Investment doubling time refers to the period it takes for an initial investment to grow twice its size when compounded at a particular interest rate. For continuous compounding, the formula \( t = \frac{\ln 2}{r} \) is used. Here, \( t \) represents time and \( r \) is the interest rate as a decimal.
The natural logarithm of 2, written as \( \ln 2 \), is approximately 0.693, which is a constant used to determine doubling time.This formula is helpful for quickly evaluating how effective an interest rate is. When the interest rate increases, the doubling time decreases. Here are the calculated doubling times for different interest rates:

• With a 6% interest rate, the doubling time is about 11.55 years.
• With a 7% interest rate, it is approximately 9.9 years.
• With an 8% interest rate, it reduces to roughly 8.66 years.

When students understand this concept, they can make informed decisions about their investments by clearly seeing the advantages of higher rates.

Exponential Growth

Exponential growth describes a process where the amount increases by a constant percentage over regular intervals. In the context of investments, this means the principal amount grows by a fixed percentage rate each period.
This growth is powerful because the increase is not just applied to the original quantity but also to its growth in previous periods. Continuous compounding is a perfect example of exponential growth. At each moment, the investment grows at an ever-increasing rate due to accrued interest. As such, even small differences in interest rates can significantly impact an investment's growth.
For students, grasping exponential growth can illuminate why investments can grow substantially faster over time and why it pays to start investing early. Understanding these mechanisms can help in personal finance and real-life applications.

Interest Rate Conversion

Converting an interest rate from percentage to decimal form is necessary for calculations involving formulas. This is because mathematical formulas, such as those for continuous compounding, typically require decimal input for accurate output.
To convert a percentage to a decimal, simply divide by 100. For instance:

• 6% becomes 0.06
• 7% converts to 0.07
• 8% translates into 0.08

These conversions allow for effective use in formulas such as \( t = \frac{\ln 2}{r} \).
Students should practice converting rates as it is a common requirement not only in compound interest problems but also in various financial and statistical computations.
Mastery of interest rate conversion ensures students accurately apply formulas, leading to correct outcomes and deeper understanding of quantitative finance principles.