Problem 107 - Calculus and its applications

Question

Solve \(P=P_{0} e^{k t}\) for \(t\).

Step-by-Step Solution

Answer

The solution for \( t \) is \( t = \frac{\ln\left( \frac{P}{P_{0}} \right)}{k} \).

Step 1: Introduce the equation

We start with the given equation:\[ P = P_{0} e^{k t} \].Our goal is to solve for \( t \).

Step 2: Isolate the exponential term

To solve for \( t \), first isolate the exponential term by dividing both sides by \( P_{0} \):\[ \frac{P}{P_{0}} = e^{k t} \].

Step 3: Apply the natural logarithm

To eliminate the exponential, take the natural logarithm (ln) of both sides of the equation:\[ \ln\left( \frac{P}{P_{0}} \right) = \ln(e^{k t}) \].Since \( \ln(e^x) = x \), the equation simplifies to:\[ \ln\left( \frac{P}{P_{0}} \right) = k t \].

Step 4: Solve for t

Now that we have \( \ln\left( \frac{P}{P_{0}} \right) = k t \), solve for \( t \) by dividing both sides by \( k \):\[ t = \frac{\ln\left( \frac{P}{P_{0}} \right)}{k} \].

Key Concepts

Natural Logarithms

Natural logarithms are logarithms with the base of the constant \( e \), approximately equal to 2.71828. They are denoted by \( \ln \). Natural logarithms are particularly useful in dealing with exponential growth or decay because they help transform exponential equations into linear forms.
For example, \( \ln(e^x) = x \). This property is crucial because it allows us to work with the exponent directly by removing the exponential nature of the equation.

Understanding \( \ln \) is essential in a variety of applications, such as in the natural sciences and finance, where exponential models are often used. If you come across an equation where a variable is in the exponent, using \( \ln \) can simplify the equation significantly, making it easier to solve.

• The natural logarithm is inverse to the exponential function.
• \( \ln(x) \) returns the power that \( e \) needs to be raised to in order to equal \( x \).
• Common use: transforming equations to make solving for unknowns easier.

Isolating Variables in Equations

Isolating a variable means arranging an equation so that one side contains only the variable you're trying to solve for. This process usually involves applying algebraic operations that "undo" the current operations applied to the variable.
In the equation \( P = P_0 e^{kt} \), isolating the term \( e^{kt} \) is a necessary step to solve for \( t \). We do this by dividing both sides by \( P_0 \), leading to \( \frac{P}{P_0} = e^{kt} \).

This technique is a fundamental aspect of algebra and can involve the following steps:

• Use inverse operations, such as adding the opposite, multiplying by reciprocals, or using logarithms, to isolate variables.
• Always perform the same operation on both sides of the equation to maintain equality.
• In more complex equations, think of it as "peeling away" layers, focusing on the outermost operation affecting the variable first.

Mastering this skill is essential for solving various algebraic equations, from the simple to the complex.

Solving for a Variable

Solving for a variable refers to the process of finding the value of an unknown that makes an equation true. This often involves manipulating the equation until the variable is isolated on one side.
For our specific problem, after isolating and simplifying using natural logarithms, we have \( \ln\left( \frac{P}{P_0} \right) = kt \). To solve for \( t \), divide both sides by \( k \), resulting in \( t = \frac{\ln\left( \frac{P}{P_0} \right)}{k} \).

Here are key points to practice when solving for variables:

• Ensure mathematical operations maintain the balance of the equation.
• Keep track of units if the equation represents a real-world scenario.
• Check the solution by substituting it back into the original equation.

This is a crucial algebra skill because it forms the basis for much of mathematical problem-solving in areas like physics, chemistry, and economics.