Question

Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0725\)

Short answer

The investment will double in approximately 9.57 years at an interest rate of 0.0725, when compounded continuously.

Step-by-step solution

Write down the given values

In this case, \(P=\$1000\), \(A=2P=\$2000\), and \(r=0.0725\). We are asked to find \(t\).

Insert the given values into the formula

Insert the given values into the formula for continuous compounding, \(A=P*e^{rt}\). This gives us \(2000 = 1000 * e^{0.0725t}\)

Solve for \(t\)

First, divide both sides of the equation by 1000. This simplifies the equation to \(2=e^{0.0725t}\). Then, take the natural logarithm (ln) of both sides to solve for \(t\). This gives us \(ln(2)=0.0725t\). Finally, divide by 0.0725 to isolate \(t\).

Compute the value

Now calculate the value of \(t\). This gives \(t = ln(2)/0.0725 \approx 9.57\) years.

Key concepts

Exponential Growth

Exponential growth is a fundamental concept often encountered in natural phenomena and financial contexts, such as the continuous compounding of interest. When an investment is subjected to exponential growth, it increases at a rate proportional to its current value. This leads to the investment's value growing faster over time as the investment itself grows.

The formula that characterizes exponential growth in continuous compounding is:

• \[ A = P \cdot e^{rt} \]

where:

• \( A \) is the amount of money accumulated after time \( t \), including interest,
• \( P \) is the principal amount (initial investment),
• \( r \) is the annual interest rate (expressed as a decimal),
• \( t \) is the time the money is invested for, and,
• \( e \) is the base of natural logarithms, approximately equal to 2.71828.

Exponential growth means that as time goes on, the amount of growth each period increases as the starting amount increases. This is very different from linear growth, where the increase is constant over time.

Natural Logarithms

Natural logarithms, denoted by \( \ln \), are logarithms that use the constant \( e \) (approximately 2.71828) as their base. In many applications in mathematics and science, natural logarithms simplify expressions and make solving logarithmic equations more intuitive.

To understand their role in continuous compounding, recognizing that the process of solving for \( t \) involves expressions of the form \( e^{x} = y \) is essential. Solving for \( x \) involves taking the natural logarithm on both sides, yielding \( x = \ln(y) \). Thus, if you have:

• \[ e^{0.0725t} = 2 \]

Taking the natural logarithm of both sides results in:

• \[ 0.0725t = \ln(2) \]

This operation highlights how natural logarithms 'undo' exponential functions, significantly simplifying equations that involve exponential terms. They are indispensable in finding the time it takes for investments to reach a desired value in problems involving continuous compounding.

Interest Rate Calculation

Calculating interest rates in financial problems involves understanding how different compounding methods impact overall growth. In continuous compounding, the formula \( A = P \cdot e^{rt} \) reveals how interest accrues smoothly over time, contributing to faster growth compared to traditional compounding methods.

In the given problem, determining the time \( t \) for the investment to double involves isolating \( t \) in:

• \[ A = P \cdot e^{rt} \]
• \[ 2 = e^{0.0725t} \]

The goal is to come up with a way to rearrange this equation to solve for a variable. Here, interest rate calculation helps establish the relationship between how fast the principal grows and the time it takes to achieve a certain amount.

Though interest rates are often annualized, they convert to a form conducive to the context of continuous growth through the calculation of \( t \) using formulas and transformations involving natural logarithms. As seen here, the calculated time \( t = \ln(2) / 0.0725 \) yields approximately 9.57 years, promising an insight into how rapidly investments can grow at a specified interest rate that compounds continuously.