Problem 24 - Calculus

Question

Solve for \(t .\) Assume \(a\) and \(b\) are positive constants and \(k\) is nonzero. $$P=P_{0} e^{k t}$$

Step-by-Step Solution

Answer

\( t = \frac{\ln \left( \frac{P}{P_0} \right)}{k} \)

Step 1: Isolate the Exponential Term

Start by dividing both sides of the equation by \( P_0 \) to isolate the exponential expression. This gives us: \[ \frac{P}{P_0} = e^{kt} \].

Step 2: Apply the Natural Logarithm

Apply the natural logarithm to both sides of the equation to eliminate the exponential term. This gives us: \[ \ln \left( \frac{P}{P_0} \right) = \ln(e^{kt}) \].

Step 3: Simplify the Equation

Use the logarithmic identity \( \ln(e^x) = x \) to simplify the right side of the equation. Thus, we have: \[ \ln \left( \frac{P}{P_0} \right) = kt \].

Step 4: Solve for \(t\)

Isolate \(t\) by dividing both sides of the equation by \(k\). This results in the final solution:\[ t = \frac{\ln \left( \frac{P}{P_0} \right)}{k} \].

Key Concepts

Logarithmic Functions

Logarithmic functions are a fundamental concept in mathematics, especially when dealing with exponential growth and decay. Essentially, a logarithm is the inverse operation of exponentiation. For instance, if you have an equation of the form \( b^x = y \), the logarithmic equivalent would be \( ext{log}_b(y) = x \). The logarithm is saying: "To what power must we raise \( b \) to get \( y \)?"
In our original exercise, we use the natural logarithm to handle the exponential equation \( P = P_0 e^{kt} \). By taking the logarithm, specifically the natural logarithm (\( ext{ln} \)), we can solve equations that involve exponentials by transforming them into a simpler linear form.
Understanding logarithmic functions helps you to:

• Convert between exponential and linear equations.
• Solve problems involving exponential growth and decay, such as interest rate calculations, population growth, and radioactive decay.
• Understand complex systems that naturally follow exponential growth patterns.

Mastering logarithmic functions is not only crucial for solving mathematical equations but also for understanding real-life phenomena.

Natural Logarithms

The natural logarithm, denoted as \( ext{ln}(x) \), is a logarithm with base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. This is the type of logarithm that appears in many natural processes, such as population growth, radioactive decay, and compound interest.
In equations, the natural logarithm offers a convenient way of simplifying expressions where the base is \( e \). For example, in \( ext{ln}(e^x) = x \), the logarithm effectively "cancels out" the exponential base \( e \), revealing the exponent. This makes natural logarithms particularly useful in solving exponential equations.
Key properties include:

• \( ext{ln}(1) = 0 \) because \( e^0 = 1 \).
• \( ext{ln}(e) = 1 \) as \( e^1 = e \).
• \( ext{ln}(e^x) = x \), which is used to simplify expressions significantly.

The natural logarithm is an essential mathematical concept that allows one to work with continuous growth or decay processes that are modelled by exponential functions.

Solving Exponential Equations

Solving exponential equations, like the one given in the original exercise, involves isolating the exponential term and then using logarithms to eliminate it. To solve these types of equations effectively, you should follow a series of steps:
1. **Isolate the Exponential Part:** Operate on the equation to get the exponential term by itself. In our exercise, that meant dividing both sides by \( P_0 \) to get \( \frac{P}{P_0} = e^{kt} \).
2. **Apply the Logarithm:** Use the natural logarithm to both sides of the equation. This simplifies the expression by removing the exponential base \( e \), resulting in \( \ln \left( \frac{P}{P_0} \right) = kt \).
3. **Solve for the Variable:** With the logarithm applied, you can isolate the variable. By dividing both sides by \( k \), you find that \( t = \frac{\ln \left( \frac{P}{P_0} \right)}{k} \).
These steps help transform an exponential equation into a manageable linear equation, making it simpler to find the unknown. Understanding these techniques is crucial for dealing with real-world situations where growth or decay rates need to be calculated.