Problem 73 - A Graphical Approach to College Algebra

Question

$$\text { Solve each formula for the indicated variable.}$$ $$A=T_{0}+C e^{-k}, \text { for } k$$

Step-by-Step Solution

Answer

\( k = -\ln\left(\frac{A - T_{0}}{C}\right) \)

Step 1: Isolate the Exponential Term

Start with the equation \( A = T_{0} + C e^{-k} \). Our goal is to solve for \( k \). First, isolate the exponential term by subtracting \( T_{0} \) from both sides: \( A - T_{0} = C e^{-k} \).

Step 2: Divide Both Sides by Constant

Next, divide both sides of the equation by \( C \) to further isolate \( e^{-k} \): \( \frac{A - T_{0}}{C} = e^{-k} \).

Step 3: Apply the Natural Logarithm

Apply the natural logarithm \( \ln \) to both sides of the equation to remove the exponent and solve for \( k \). We have: \( \ln\left(\frac{A - T_{0}}{C}\right) = -k \).

Step 4: Solve for k

To finish solving for \( k \), multiply both sides by -1: \( k = -\ln\left(\frac{A - T_{0}}{C}\right) \). This is the expression for \( k \) in terms of \( A \), \( T_{0} \), and \( C \).

Key Concepts

Exponential Equations

Exponential equations include variables in the exponent. This makes them unique and slightly tricky compared to linear equations. In this exercise, the equation \( A = T_0 + Ce^{-k} \) is an example of such an equation.
Here, \( e^{-k} \) is an exponential term where \( e \) is the base of natural logarithms. Exponential equations are common in real-world problems involving growth and decay, like population growth or radioactive decay.
To solve exponential equations, focus on these steps:

• Identify the exponential term: \( e^{-k} \) in our case.
• Isolate the exponential part of the equation using basic algebraic operations.

This isolation is crucial because it sets you up for the next step, which involves logarithms.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It is essential for solving equations where the variable is in the exponent.
In this exercise, after isolating the exponential term, we have \( e^{-k} \) equal to some expression. To 'solve' this exponent, we apply the natural logarithm. This operation helps to bring down the exponent into a more manageable form.
Here's why natural logarithms are useful:

• The natural log of \( e^x \) is simply \( x \), i.e., \( \ln(e^x) = x \).
• This property allows us to rewrite the problem in a linear form, making it easier to solve for the variable \( k \).

So when you see \( \ln\left(\frac{A - T_0}{C}\right) = -k \), you can easily find \( k \) by rearranging the terms.

Isolating Variables

Isolating variables is a foundational technique in algebra that allows us to solve equations. It involves manipulating the equation until the variable of interest is by itself on one side of the equation.
In solving for \( k \) from the equation \( A = T_0 + Ce^{-k} \), isolating the exponential component was the first key step. This set the stage to turn the exponential equation into a solvable form by using logarithms.
Here’s a brief guide on isolating variables:

• Perform operations to clear out any constants mixed with the variable.
• Each step should bring you closer to having the variable alone on one side of the equation.
• Once the variable is isolated, apply any necessary functions like logarithms to simplify further.

This process transforms challenging equations into solvable forms. In our case, it made finding \( k = -\ln\left(\frac{A - T_0}{C}\right) \) possible and straightforward.