Problem 63 - Precalculus : Building Concepts and Connections

Question

Determine how long it takes for the given investment to double if \(r\) is the interest rate and the interest is compounded continuously. Assume that no withdrawals or further deposits are made. Initial amount: \(\$ 4000 ; r=5.75 \%\)

Step-by-Step Solution

Answer

It will take approximately 12.06 years for the investment to double.

Step 1: Identify the given parameters and the formula to be used

The initial investment (P) is given as $4000 and the annual interest rate (r) is 5.75% or 0.0575 in decimal. We are trying to find the time (t) it takes for the investment to double. So, we set A = 2P = $8000. The formula for continuous compound interest is \( A = Pe^{rt} \).

Step 2: Input the values into the formula

Input the given values into the formula: \( 8000 = 4000 \times e^{0.0575t} \).

Step 3: Simplify the equation

Firstly, divide both sides by 4000, resulting in: \(2 = e^{0.0575t} \). Secondly, to solve for \(t\), take the natural log (ln) of both sides: \( \ln 2 = \ln e^{0.0575t} \).

Step 4: Solve for \(t\)

We can simplify \( \ln e^{0.0575t} \) to \( 0.0575t \). So, \( \ln 2 = 0.0575t \). The equation now becomes \(t = \frac{ \ln 2 }{0.0575}\).

Step 5: Evaluate the final answer

This gives us \( t ≈ 12.06 \) years.

Key Concepts

Investment Doubling Time

Investment doubling time refers to the period it takes for an initial investment to double in value, under a specific interest rate and compounding method. When dealing with continuous compounding, as in our exercise, this means interest is applied at every moment in time, which allows for continuous growth. Here's a simple way to think about it:

• You have an initial sum of money.
• You want to know how long before it's twice as much.
• No extra money is added except the earned interest.

To calculate this doubling time, we use the formula for continuous compound interest: \[ A = P e^{rt} \]where

• \( A \) = the final amount,
• \( P \) = the initial principal balance,
• \( r \) = the rate of interest per annum,
• \( t \) = the time the money is invested for in years.

In the exercise, setting \( A \) to 2P or double the initial amount helps us determine how long it will take for the investment to reach twice its value.

Natural Logarithms

Natural logarithms (often written as \( \ln \)) are logarithms with base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. They are particularly useful in continuous growth calculations, like those used in continuous compound interest.In the context of our exercise, solving the compound interest equation involves taking the natural logarithm:\[ \ln 2 = \ln e^{0.0575t} \]By applying logarithm rules, we can simplify \( \ln e^{0.0575t} \) to the exponent itself, i.e., \( 0.0575t \). This simplification is possible because the \( \ln \) and \( e^{\cdot} \) functions are inverses of one another. This helps us isolate the variable t to solve for the time it takes for the investment to double.

Exponential Growth

Exponential growth occurs when a quantity increases by a consistent percentage over time. The growth rate in an exponential function determines how fast this quantity grows, which is pertinent when discussing investments and compounding interest.With continuous compounding, this growth is theoretically occurring at every instant. The formula \( A = Pe^{rt} \) describes how the investment amount A grows exponentially over time. Here's what happens:

• The principal amount \( P \), such as \$4000 in the exercise, grows gradually.
• The rate \( r \) affects how quickly this growth happens. In our exercise, it's 5.75% expressed as a decimal, 0.0575.
• Time \( t \) is the duration for which this growth is observed, influenced by \( r \).

Exponential growth shows how rapidly an investment can increase with the power of compounding, often leading to more significant results than simple interest over extended periods.