Problem 34 - Intermediate Algebra

Question

How long will it take $$\$ 1000$$ to double itself if it is invested at \(5 \%\) interest compounded semiannually?

Step-by-Step Solution

Answer

It will take approximately 14.03 years.

Step 1: Understand the Formula

The formula for compound interest is \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where \( A \) is the future value of the investment/loan, including interest, \( P \) is the principal investment amount (the initial deposit or loan amount), \( r \) is the annual interest rate (decimal), \( n \) is the number of times that interest is compounded per year, and \( t \) is the number of years the money is invested for.

Step 2: Set Up the Equation

To find out when the investment doubles, \( A \) should be \( 2P \). Here, \( P = 1000 \). Thus, \( 2P = 2000 \). Use this in the formula: \( 2000 = 1000 \left(1 + \frac{0.05}{2}\right)^{2t} \).

Step 3: Simplify the Equation

Divide both sides by \( 1000 \) to isolate the compound interest term: \( 2 = \left(1 + \frac{0.05}{2}\right)^{2t} \). Simplify \( 1 + \frac{0.05}{2} = 1 + 0.025 = 1.025 \). Thus, the equation becomes \( 2 = 1.025^{2t} \).

Step 4: Solve for Time (t)

To solve for \( t \), take the natural logarithm of both sides: \( \ln(2) = \ln(1.025^{2t}) \). Using logarithmic identities, this becomes \( \ln(2) = 2t \cdot \ln(1.025) \).

Step 5: Calculate t

Rearrange the formula to solve for \( t \): \( t = \frac{\ln(2)}{2 \cdot \ln(1.025)} \). Now calculate \( \ln(2) \approx 0.6931 \) and \( \ln(1.025) \approx 0.0247 \). Substitute these into the equation: \( t = \frac{0.6931}{2 \cdot 0.0247} \approx \frac{0.6931}{0.0494} \approx 14.03 \) years.

Key Concepts

Investment Doubling Time

The concept of investment doubling time refers to how long it takes for an investment to grow to twice its original amount. This is an important aspect of financial planning as it gives investors insight into how quickly their money will grow.
To determine the doubling time, investors often rely on the formula for compound interest. The critical point here is that the investment grows at a rate defined by the interest, which compounds at regular intervals. In this case, the interest is compounded semiannually.
When you set up your equation for doubling time, you replace the final amount, \( A \), with \( 2P \) because you want the final amount to be twice the principal \( P \). Simplifying the equation allows us to focus on the relationship between the rate of growth and the time it takes to reach this doubled amount.

Interest Calculation

Interest calculation in the context of compound interest involves understanding how the initial investment amount grows over time. The formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \) is central to this concept.
Each part of this formula represents a key component of how interest is calculated:

• \( P \) is the principal amount, the initial investment.
• \( r \) is the annual interest rate, expressed as a decimal.
• \( n \) is the number of times interest is compounded per year.
• \( t \) is the number of years for which the money is invested.

To see how interest affects investment growth, consider how changing the compounding frequency or interest rate alters \( A \). For instance, even with the same annual rate, an investment compounded semiannually grows differently than an investment compounded annually. This calculation allows investors to plan and predict their financial growth effectively.

Logarithmic Identities

Logarithmic identities play a crucial role when solving equations involving exponential growth, such as compound interest. Logs are used to simplify equations and solve for unknown variables effectively.
In our given problem, after simplifying the compound interest equation, it becomes necessary to solve for \( t \). This requires taking the natural logarithm of both sides, converting the exponential equation into a linear form: \( \ln(2) = 2t \cdot \ln(1.025) \) This application uses particular identities like \( \ln(a^b) = b \cdot \ln(a) \), which helps in removing the exponent so we can solve for \( t \).
Understanding these identities allows students to follow the logical steps in solving such problems and apply similar techniques to other equations requiring the calculation of time or rate in exponential conditions.