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The natural logarithm is a mathematical function often denoted as \( \ln(x) \). It is the inverse function of the exponential function with base \( e \). The number \( e \) is an irrational number approximately equal to 2.718. When we want to "undo" or "reverse" an exponential function involving \( e \), we apply the natural logarithm.

• For example, if \( e^x = a \), then \( x = \ln(a) \).

Using \( \ln \) is extremely handy when solving equations involving exponential functions, especially when trying to isolate the variable in the exponent. In the inverse function problem given, we see how the natural logarithm helps us find \( x \) after isolating \( e^x \). The step involved converting the expression \( e^x = \frac{y + 1}{1 - y} \) to \( x = \ln\left(\frac{y + 1}{1 - y}\right) \). This demonstrates the role of the natural logarithm as a tool for solving exponential equations.

Exponential functions are mathematical expressions in which a constant base—often \( e \), the mathematical constant—is raised to a variable exponent. They are written in the form \( a^x \), where \( a \) is a positive constant. Particularly, \( e^x \) is a common exponential function due to its unique properties and frequent occurrence in mathematical modeling and natural processes.
Exponential functions increase rapidly, which makes them very different from linear or polynomial functions. In our original exercise, the function \( f(x) = \frac{e^x - 1}{e^x + 1} \) involves \( e^x \), an exponential function.

• The function utilizes \( e^x \) both in the numerator and denominator, which complicates direct solving without manipulation.

This type of function often appears in problems involving growth, decay, and interest—reflecting how exponential growth can quickly escalate values. The rapid growth feature of exponential functions is mirrored in the behavior of the original function as \( x \) changes.

Solving equations involves finding the value of the variable that makes the equation true. In the context of finding inverse functions, our task is to express the input variable in terms of the output variable. This requires manipulating the equation systematically.
The given exercise walks through this process, step by step, till we isolate \( x \) using \( y \).

• Begin by setting \( y = \frac{e^x - 1}{e^x + 1} \), aligning the function in terms of \( y \).
• Clear fractions via cross-multiplication, leading to \( y(e^x + 1) = e^x - 1 \).
• Distribute and rearrange terms to group \( e^x \).
• Factor \( e^x \) and isolate it by dividing one side by \( (y - 1) \).

Finally, the natural logarithm comes into play to solve for \( x \) after expressing \( e^x \) in a simplified form. Each of these steps demonstrates the importance of a structured approach in solving equations, particularly when involving inverse functions.