Question

Use \(y=y_{0} e^{k t}\). If \(y=100\) at \(t=4\) and \(y=10\) at \(t=8\), when does \(y=1\) ?

Short answer

\( t = 12 \)

Step-by-step solution

Set up the given equations

We have two conditions given: \( y = 100 \) at \( t = 4 \) and \( y = 10 \) at \( t = 8 \). These can be represented by the equations \( 100 = y_0 e^{4k} \) and \( 10 = y_0 e^{8k} \).

Solve for initial value \(y_0\)

Divide the second equation by the first equation to eliminate \(y_0\): \( \frac{10}{100} = \frac{y_0 e^{8k}}{y_0 e^{4k}} \). This simplifies to \( \frac{1}{10} = e^{4k} \).

Solve for \(k\)

Take the natural logarithm of both sides of \( \frac{1}{10} = e^{4k} \) to solve for \(k\). So, \( ln \left( \frac{1}{10} \right) = 4k \). Simplifying gives \( k = \frac{\ln \left( \frac{1}{10} \right)}{4} \).

Use the value of \(k\) in one of the original equations

Substitute \(k\) back into one of the original equations, say \( 100 = y_0 e^{4k} \), to find \(y_0\). Use \(k = \frac{\ln \left( \frac{1}{10} \right)}{4} \) and solve for \(y_0\). Then, \( y_0 = 100 e^{-4k} = 100 \times 10 = 1000 \).

Find the time when \(y = 1\)

Now, substitute \(y_0 = 1000\) and \(k = \frac{\ln \left( \frac{1}{10} \right)}{4} \) into \( y = y_0 e^{kt} \) and set \( y = 1 \): \( 1 = 1000 e^{kt} \). Simplifying gives \( e^{kt} = \frac{1}{1000} \).

Solve for \(t\)

Take the natural logarithm of both sides to solve for \(t\): \( ln \left( \frac{1}{1000} \right) = kt \). Substitute \( k = \frac{\ln \left( \frac{1}{10} \right)}{4} \) from **Step 3**. Therefore, \( t = \frac{\ln \left( \frac{1}{1000} \right)}{k} \). Substitute \(k\) and solve for \(t\). This gives \( t = 12 \).

Key concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. This type of logarithm is widely used in mathematical models for continuous growth or decay. In exponential functions, the natural logarithm is essential for solving equations where the variable is in the exponent.

For example, if we have an equation like \( e^{4k} = \frac{1}{10} \), taking the natural logarithm on both sides helps bring the exponent down. This process enables us to solve for \( k \) by converting the equation to \( \ln \left( e^{4k} \right) = \ln \left( \frac{1}{10} \right) \).

• The natural logarithm of \( e^x \) simplifies to \( x \), making it a powerful tool in solving exponential equations.
• Using \( \ln \) helps isolate and solve for variables that are part of an exponent.

Solving Exponential Equations

Solving exponential equations involves finding the unknown variable in an equation where the variable appears in the exponent. The general approach is to manipulate the equation so that one side is the exponentiated result, and then use logarithms to remove the variable from the exponent.

In our exercise, we used the equation \( y = y_0 e^{kt} \) representing exponential growth or decay. Exponential growth occurs when the constant \( k \) is positive, and decay occurs if \( k \) is negative. By solving equations such as \( 100 = y_0 e^{4k} \) and \( 10 = y_0 e^{8k} \), we determine crucial parameters like the initial value \( y_0 \) and the growth/decay rate \( k \).

• Start by setting up relevant equations based on given data points such as \( y(t_1) \) and \( y(t_2) \).
• Eliminate constants by dividing the equations to simplify terms, which often reveals hidden relationships like \( e^{4k} = \frac{1}{10} \).
• Use logarithms to solve for the component in the exponent, crucial in determining \( k \).

Initial Value Problem

An initial value problem entails determining an unknown constant or function in relation to an initial condition. In the context of exponential functions, the 'initial value' refers to the value of \( y_0 \) when \( t = 0 \).

In our exercise, finding \( y_0 \) required solving equations that provided conditions at different times (\( t = 4 \) and \( t = 8 \)). By manipulating these conditions, such as through division, we eliminate mathematical variables to isolate our desired unknowns like \( y_0 \).

Initial value problems are common in scenarios involving growth and decay, such as population studies or radioactive decay. Understanding the initial state gives profound insights into the entire process.

• Identify initial conditions from the problem to set up equations.
• Use known values and solved constants like \( k \) to find \( y_0 \).
• Solutions lead to a complete function representing the system's behavior over time.