Problem 75 - A Graphical Approach to Precalculus with Limits

Question

Solve each formula for the indicated variable. $$y=A+B\left(1-e^{-C x}\right), \text { for } x$$

Step-by-Step Solution

Answer

\( x = -\frac{1}{C} \ln\left(1 - \frac{y - A}{B}\right) \)

Step 1: Isolate the Exponential Term

Start with the equation \( y = A + B(1 - e^{-Cx}) \). Subtract \( A \) from both sides to isolate the exponential term: \( y - A = B(1 - e^{-Cx}) \).

Step 2: Divide by B

Divide both sides by \( B \) to further isolate the exponential term: \( \frac{y - A}{B} = 1 - e^{-Cx} \).

Step 3: Solve for the Exponential Term

Rearrange the equation to solve for the exponential term by subtracting 1 from both sides: \( e^{-Cx} = 1 - \frac{y - A}{B} \).

Step 4: Natural Logarithm Elimination

Take the natural logarithm of both sides to eliminate the exponential function: \( -Cx = \ln\left(1 - \frac{y - A}{B}\right) \).

Step 5: Isolate x

Solve for \( x \) by dividing both sides by \( -C \): \( x = -\frac{1}{C} \ln\left(1 - \frac{y - A}{B}\right) \).

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in precalculus and are pivotal in numerous applications across science and engineering. These functions are expressed in the form \( f(x) = a \, b^{x} \), where \( a \) is a constant, \( b \) is the base of the exponential (a positive number not equal to 1), and \( x \) is the exponent.

In the context of the given exercise, the function \( y = A + B(1 - e^{-Cx}) \) is examined, where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. The formula represents a type of exponential decay or growth, depending on the sign of \( C \). When \( C \) is positive, \( e^{-Cx} \) describes a decay, as the exponent \( -Cx \) leads to division by progressively larger sums, reducing \( e^{-Cx} \) rapidly to zero.

Understanding exponential functions involves:

• Identifying the base and the exponent
• Recognizing the role of the constant \( e \) in natural exponential functions
• Exploring the applications in modeling real-world phenomena like population growth and radioactive decay

Grasping these elements not only assists in mathematical computations but also in interpreting how quantities change over time.

Natural Logarithms

Natural logarithms are logarithms with the base \( e \), denoted as \( \ln(x) \). They provide powerful tools for solving equations involving exponential terms. The key property of logarithms, that they are the inverse of exponential functions, makes them particularly useful. For instance, the equation \( y = A + B(1 - e^{-Cx}) \) becomes manageable once simplified using natural logs.

When we need to "undo" an exponential function, we apply the natural logarithm. In the given exercise, taking the natural log of both sides of the equation isolates the variable \( x \). This is achieved because \( \ln(e^{-Cx}) = -Cx \), thanks to the property that \( \ln(e^x) = x \).

Important aspects of natural logarithms include:

• Their ability to convert complex multiplication into simple addition
• The intrinsic relationship \( e^{\ln(x)} = x \)
• Applications in financial mathematics for continuous compounding

Understanding natural logarithms is crucial for both theoretical mathematics and practical problem-solving.

Equation Solving

Equation solving is a fundamental skill in mathematics that transcends simple arithmetic toward more complex algebraic manipulations. Solving equations, particularly those involving exponential or logarithmic terms, necessitates a logical approach to isolate the desired variable.

In the exercise at hand, our goal is to solve for \( x \) in the equation \( y = A + B(1 - e^{-Cx}) \).

Here's a step-by-step breakdown of the process used in this solution:

• Initially isolate the exponential expression by manipulating the equation, i.e., subtracting constants and dividing by coefficients.
• Next, apply the natural logarithm to both sides, leveraging its property as the inverse of the exponential function, to eliminate the exponential component.
• Lastly, perform algebraic manipulation to solve for \( x \), usually by dividing or multiplying both sides of the equation by any remaining coefficients or constants.

These steps highlight the balance of operations needed to isolate and solve for a specific variable, particularly common in solving equations derived from real-world problems.