Problem 53 - Algebra 2

Question

A savings account earns 4.62\(\%\) annual interest, compounded continuously. After approximately how many years will a principal of \(\$ 500\) double? \(\begin{array}{llll}{\text { A. } 2 \text { years }} & {\text { B. } 10 \text { years }} & {\text { C. } 15 \text { years }} & {\text { D. } 44 \text { years }}\end{array}\)

Step-by-Step Solution

Answer

Calculating \(t = \frac{ln(2)}{0.0462}\) gives us approximately 15 years, so the correct option is C: 15 years.

Step 1: Setting up the Equation

First, insert the given values into the continuous compound interest formula \(A = Pe^{rt}\). This gives us \(1000 = 500e^{0.0462t}\).

Step 2: Simplifying the Equation

Next, the equation can be simplified by dividing both sides by 500. Thus, we have \(2 = e^{0.0462t}\).

Step 3: Solving for t

To solve for \(t\), we take the natural logarithm (ln) on both sides, which gives us \(ln(2) = 0.0462t\). Hence, \(t = \frac{ln(2)}{0.0462}\).

Key Concepts

Exponential Growth

When we talk about exponential growth, we mean a process that grows at a rate proportional to its current size. In mathematical terms, it's often represented with functions involving the base of the natural logarithm, denoted as \(e^{rt}\).
This is particularly relevant in continuous compounding, where interest is added to the principal in such a manner that it seems to be compounded at every possible moment. This results in an exponential increase in the amount over time.
For example, consider investing \(\$500\) with interest compounded continuously at the rate of 4.62%. To find when this investment doubles, we use the concept of exponential growth. Here, \(P\) represents the initial principal, \(e\) is a constant approximately equal to 2.71828, \(r\) is the annual interest rate, and \(t\) is the time in years.
Remember, the key feature of exponential growth is how quickly the amount grows, which is why even small interest rates over long periods can yield substantial returns.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is simply the logarithm to the base \(e\). It helps in solving equations where the exponent is unknown, as seen in exponential growth scenarios.
For continuous compound interest problems, using the natural logarithm becomes essential. When you have an equation like \(2 = e^{0.0462t}\), to solve for \(t\), you use \(\ln\) to rewrite it as \(\ln(2) = 0.0462t\). This turns the exponential equation into a linear one, making it manageable.
Here's a quick tip: the natural logarithm will always need to use "\(e\)" as the base. What really simplifies your life is knowing that \(\ln(e^x) = x\). This property allows you to eliminate the exponential component and solve for \(t\) directly. That's also why when doubling the principal amount, you end up using \(\ln(2)\), simplifying the process significantly.

Financial Mathematics

Financial mathematics combines mathematical methods to solve problems and model market behaviors in finance. It's vital for topics like interest calculations, investments, and savings.
Continuous compounding is a frequent topic, where the formula \(A = Pe^{rt}\) is employed to calculate future investment values. Let's break this formula down a bit:

• \(A\) is the future amount of money, including interest.
• \(P\) is the initial principal or starting amount invested.
• \(r\) is the annual nominal interest rate (as a decimal).
• \(t\) is the time the money is invested for, in years.

This formula is integral to financial mathematics as it enables precise future value calculations when dealing with compounding interests. Knowing how to manipulate it by using logarithms, as seen earlier, empowers you to determine periods needed to achieve specific financial goals, like doubling an investment.
The beauty of financial mathematics is how it provides a structured way to analyze investments and savings, offering strategic insights for making sound financial decisions.