Problem 74 - A Graphical Approach to College Algebra

Question

$$\text { Solve each formula for the indicated variable.}$$ $$y=\frac{K}{1+a e^{-b x}}, \text { for } b$$

Step-by-Step Solution

Answer

\(b = -\frac{1}{x} \ln\left(\frac{K-y}{ya}\right)\)

Step 1: Isolate the Denominator

The given formula is \( y = \frac{K}{1+ae^{-bx}} \). To solve for \( b \), we first aim to isolate the denominator containing \( b \). Multiply both sides by \( 1 + ae^{-bx} \) to get: \( y(1 + ae^{-bx}) = K \).

Step 2: Distribute and Rearrange

Distribute \( y \) on the left side to get: \( y + yae^{-bx} = K \). Rearrange to isolate the exponential term: \( yae^{-bx} = K - y \).

Step 3: Solve for the Exponential Term

Divide both sides by \( ya \) to solve for the exponential component: \( e^{-bx} = \frac{K-y}{ya} \).

Step 4: Take the Natural Logarithm

Apply the natural logarithm (\( \ln \)) to both sides to eliminate the exponent: \( -bx = \ln\left(\frac{K-y}{ya}\right) \).

Step 5: Solve for b

Divide both sides by \(-x\) to isolate \( b \): \( b = -\frac{1}{x} \ln\left(\frac{K-y}{ya}\right) \).

Key Concepts

Solving Equations

When solving equations, the main goal is to find the value of the unknown variable. This often involves various algebraic manipulations such as:

• Adding or subtracting the same value from both sides of the equation
• Multiplying or dividing both sides by the same non-zero value
• Rearranging terms to isolate the desired variable

In our exercise, we wanted to solve for the variable \( b \). Initially, \( b \) was embedded in the exponent, so our first task was isolating the term where \( b \) appears. By multiplying both sides by \( 1 + ae^{-bx} \), we successfully moved the denominator to the other side.
Next, distributing the \( y \) allowed us to identify the exponential term that includes \( b \). Through these careful manipulations, the equation evolves systematically, and we gradually isolate our variable of interest.

Exponential Functions

Exponential functions include any expression where the variable is an exponent. They typically appear in the form of \( a^x \), where \( a \) is a constant and \( x \) is the variable. These functions exhibit rapid growth or decay, depending on the base.
In the context of our problem, the term \( e^{-bx} \) is an example of an exponential function. Here, \( e \) represents Euler's number, approximately 2.718, and \( b \) is the variable in question.
By nature, exponential functions can be tricky because variables in exponents can make solving equations challenging. To deal with them effectively, it often helps to employ logarithms, which can "bring down" the exponent as a coefficient, making the variable easier to extract from the exponent.

Natural Logarithm

Natural logarithms, denoted as \( \ln \), are logarithms with base \( e \). They are particularly handy for solving equations involving the exponential function \( e^x \).
When we encounter exponential terms like \( e^{-bx} \), applying the natural logarithm can simplify things. By using \( \ln \), the exponential term turns into a linear one: \( \ln(e^{-bx}) = -bx \cdot \ln(e) = -bx \) because \( \ln(e) = 1 \).
This transformation is pivotal when seeking unknowns in exponents. In our problem, after isolating \( e^{-bx} \), we applied the natural logarithm to both sides of the equation. This conversion allowed us to finally express \( b \) as a simple fraction: \( b = -\frac{1}{x} \ln\left(\frac{K-y}{ya}\right) \), neatly displaying \( b \) as the subject.