Question

Use \(y=y_{0} e^{k t}\). Suppose the value of \(\$ 1\) in Japanese yen decreases at \(2 \%\) per year. Starting from \(\$ 1=¥ 250\), when will \(\$ 1=¥ 1 ?\)

Short answer

It will take approximately 279.39 years for \$1 to decrease from ¥250 to ¥1.

Step-by-step solution

Understand the equation

We are using the exponential decay formula: \( y = y_0 e^{kt} \), where \( y \) is the amount at time \( t \), \( y_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time.

Identify the known values

In this problem, the initial value \( y_0 \) is 250 yen (\( y_0 = 250 \)), the final value \( y \) is 1 yen (\( y = 1 \)), and the rate of decrease is 2% per year. This means the decay constant \( k \) is \(-0.02\).

Set up the equation

Substitute the known values into the exponential decay equation: \( 1 = 250 e^{-0.02t} \).

Solve for time \( t \)

First, isolate the exponential term: \( e^{-0.02t} = \frac{1}{250} \). Then, take the natural logarithm (ln) of both sides to solve for \( t \): \( -0.02t = \ln(\frac{1}{250}) \).

Calculate the time \( t \)

Calculate \( t \) using the equation: \( t = \frac{\ln(\frac{1}{250})}{-0.02} \). This evaluates to approximately \( t = 279.39 \) years.

Key concepts

Exponential Growth and Decay

Exponential growth and decay are concepts used to model situations where a quantity increases or decreases at a rate proportional to its current value. This change is often depicted using an exponential function. When a quantity grows, such as population or investment, it increases over time. Conversely, when a quantity decays, like radioactive substances or the value of currency, it decreases over time.

In exponential decay, the equation is often represented as \( y = y_0 e^{kt} \), where:

• \( y \) is the final amount
• \( y_0 \) is the initial amount
• \( k \) is the decay constant (negative for decay)
• \( t \) is time

The decay constant, \( k \), is crucial as it determines the rate at which the quantity decreases. A negative \( k \) signifies decay, where a small negative value indicates slow decay, while a larger negative value indicates faster decay.

In the given exercise, the value of one dollar in yen is decreasing by 2% per year, a classic example of exponential decay. The initial yen value is 250, and we're interested in finding when it reaches 1 yen.

Natural Logarithms

Natural logarithms are a key tool in solving exponential decay problems. The natural logarithm, denoted by \( \ln \), is the inverse function of the exponential function \( e^x \).

When solving for time in exponential decay, we often take the natural logarithm of both sides of the equation. For instance, if you have an equation \( e^{kt} = a \), applying the natural logarithm will simplify this to \( kt = \ln(a) \).

This step allows us to solve for \( t \), the variable representing time. In the context of the exercise, we transformed the equation \( e^{-0.02t} = \frac{1}{250} \) to \( -0.02t = \ln(\frac{1}{250}) \).

This application of natural logarithms is essential to unraveling exponential equations, making it simpler to calculate the duration needed for currency conversion changes.

Mathematical Modeling

Mathematical modeling involves using mathematics to represent, analyze, make predictions, or otherwise provide insight into real-world phenomena. In financial contexts, models often involve exponential growth or decay to predict values over time.

Through mathematical modeling, we can gain significant insights into various processes, such as currency devaluation or economic forecasts. In the exercise, we modeled the devaluation of a dollar to yen using the exponential decay function. We chose parameters like the initial value and decay rate to set up an accurate model.

By substituting the known values into the exponential equation, we predict the time it takes for a dollar value in yen to decrease from 250 to 1. This kind of model helps individuals and businesses plan for future financial scenarios by understanding trends and allowing for better decision-making.

Therefore, mathematical modeling not only provides a snapshot of current trends but also serves as a powerful predictive tool for future events.