Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-42: Use natural logarithms to solve for \(x\) in terms of \(y\). $$ y=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} $$

Step-by-Step Solution

Answer

\( x = \frac{1}{2}\ln\left(\frac{1+y}{y-1}\right) \) in terms of \( y \).

Step 1: Express the equation in terms of exponentials

The given equation is \( y = \frac{e^x + e^{-x}}{e^x - e^{-x}} \). We need to solve for \( x \) in terms of \( y \).

Step 2: Clear the fraction by cross-multiplying

Multiply both sides by \( e^x - e^{-x} \) to eliminate the denominator on the right side:\[ y(e^x - e^{-x}) = e^x + e^{-x} \]

Step 3: Distribute \( y \)

Distribute \( y \) on the left side:\[ ye^x - ye^{-x} = e^x + e^{-x} \]

Step 4: Rearrange the equation

Rearrange terms to bring all \( e^x \) terms to one side and \( e^{-x} \) terms to the other side:\[ ye^x - e^x = e^x + e^{-x} + ye^{-x} \]

Step 5: Factor the equation

Factoring out the common terms, we get:\[ e^x(y - 1) = e^{-x}(1 + y) \]

Step 6: Express in terms of exponentials again

Since \( e^{-x} = \frac{1}{e^x} \), rewrite the equation:\[ e^{2x}(y - 1) = (1 + y) \]

Step 7: Isolate \( e^{2x} \)

Solve for \( e^{2x} \):\[ e^{2x} = \frac{1 + y}{y - 1} \]

Step 8: Take the natural logarithm of both sides

Apply the natural logarithm to both sides to solve for \( x \):\[ 2x = \ln\left(\frac{1 + y}{y - 1}\right) \]

Step 9: Solve for \( x \)

Divide by 2 on both sides to solve for \( x \):\[ x = \frac{1}{2}\ln\left(\frac{1 + y}{y - 1}\right) \]

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are fundamental because they model various real-world scenarios like population growth, compound interest, and radioactive decay. In particular, the base of natural exponential functions is the mathematical constant \( e \), approximately equal to 2.71828.Understanding how these functions behave is crucial:

• If the exponent is positive, the function increases (e.g., \( e^x \)).
• If the exponent is negative, the function decreases (e.g., \( e^{-x} \)).

Exponential functions have unique properties that make them easy to manipulate in algebraic contexts. For instance, they allow us to transform complex multiplication into simpler addition, a process frequently useful in solving equations.

Cross-Multiplication

Cross-multiplication is a strategy used to solve equations where a variable is part of a ratio or fraction. It involves multiplying the numerator of one side by the denominator of the other and vice versa. This technique is handy for eliminating fractions, simplifying the process of solving more complex equations.Let's consider an example using the initial equation:
Given \( y = \frac{e^x + e^{-x}}{e^x - e^{-x}} \), cross-multiply to eliminate the fraction:

• Multiply \( y \) by \( e^x - e^{-x} \).
• Multiply \( 1 \) by \( e^x + e^{-x} \).

The resulting equation, \( y(e^x - e^{-x}) = e^x + e^{-x} \), no longer contains fractions on either side, making it easier to solve for \( x \). By applying cross-multiplication, we transform a rational equation into a polynomial form, simplifying further analysis.

Solving Equations

Solving equations involves finding the value(s) of variables that make the equation true. When dealing with exponential equations like \( e^{2x} = \frac{1+y}{y-1} \), the process requires specific steps:1. **Isolate the exponential term:** Rearrange the equation such that \( e^{2x} \) stands alone.2. **Apply logarithms:** Use natural logarithms (\( \ln \)) to remove the exponential component. Logarithms allow us to convert multiplication relationships into additive ones.3. **Solve for the variable:** Further manipulate the equation algebraically to find \( x \) in terms of other variables.With our example, applying the natural logarithm to both sides yields \( 2x = \ln\left(\frac{1+y}{y-1}\right) \). This crucial step simplifies the exponential function, allowing for straightforward algebraic manipulation to solve for \( x \).

Algebraic Manipulation

Algebraic manipulation encompasses a variety of techniques used to rearrange and simplify equations. Key strategies include factoring, distributing, and combining like terms. Mastery of algebraic manipulation is crucial in solving complex equations effectively.In our given scenario, algebraic manipulation occurs throughout the solution process:

• **Distributing** \( y \) on both sides of the equation to simplify terms.
• **Factoring** the equation by identifying common terms \( (e^x)(y-1) = (e^{-x})(1+y) \).
• **Using identities** such as \( e^{-x} = \frac{1}{e^x} \) to transform the equation \( e^{2x}(y-1) = 1+y \).
• **Isolating** variables to progress towards solving for \( x \): \( e^{2x} = \frac{1+y}{y-1} \).

These steps highlight how each algebraic technique simplifies and modifies the original equation, ultimately leading to the solution \( x = \frac{1}{2}\ln\left(\frac{1+y}{y-1}\right) \).