Problem 81 - Beginning and Intermediate Algebra

Question

Solve using any method. $$\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=3$$

Step-by-Step Solution

Answer

The short answer based on the given step-by-step solution is: \(x=\ln(2)\).

Step 1: Perform a substitution

To simplify the equation, let's make a new variable substitution. We will let \(y=e^{x}\), so \(e^{-x}=\frac{1}{y}\). Our equation will now be in terms of y: \(\frac{y+\frac{1}{y}}{y-\frac{1}{y}}=3\)

Step 2: Clear the denominator

In order to remove the fractions from our equation, multiply both sides by the denominator (\(y-\frac{1}{y}\)): \((y+\frac{1}{y})(y-\frac{1}{y})=3(y-\frac{1}{y})\)

Step 3: Simplify the equation

Distribute the left side of the equation and simplify: \(y^2-1=3y-3\) Now, move all terms to the left side and simplify further: \(y^2-3y+2=0\)

Step 4: Factor the quadratic equation

Factor the equation to find the factors of y: \((y-2)(y-1)=0\)

Step 5: Solve for y

Set each factor to zero and solve for y: \(y-2=0 \Rightarrow y=2\) and \(y-1=0 \Rightarrow y=1\)

Step 6: Convert back to x

Recall that we let \(y=e^{x}\). Therefore, we will now convert our solutions for y back to x: \(e^{x}=2\) and \(e^{x}=1\) Solve for x: \(x=\ln(2)\) and \(x=\ln(1)\)

Step 7: Verify the solution

Check if the values of x satisfy the original equation: \[\frac{e^{\ln(2)}+e^{-\ln(2)}}{e^{\ln(2)}-e^{-\ln(2)}}=3\] This equation simplifies to \(\frac{2+\frac{1}{2}}{2-\frac{1}{2}}\), which equals 3, so the first value of x is correct. Now check the second value: \[\frac{e^{\ln(1)}+e^{-\ln(1)}}{e^{\ln(1)}-e^{-\ln(1)}}=\frac{1+1}{1-1}\] This equation simplifies to \(\frac{2}{0}\), which is undefined. So, the second value of x is not a valid solution. So, the only solution to the given equation is \(x=\ln(2)\).

Key Concepts

Substitution Method

When solving equations, it is often helpful to simplify the problem by using the substitution method. Substituting means replacing a certain part of the equation with a new variable to make it easier to work with.
For instance, in the given problem where the expression involves terms like \( e^{x} \) and \( e^{-x} \), we use substitution by letting \( y = e^{x} \). This then transforms \( e^{-x} \) to \( \frac{1}{y} \). By making this substitution, the complex exponential equation becomes a simpler rational equation, allowing us to focus on solving for \( y \).

• Substitution helps to reorganize and simplify equations.
• It provides a clearer path by reducing complex terms.

Remember that once the new variable is solved, you must convert back to the original variable to find the final solution to the problem.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). They are solved using several methods such as factoring, using the quadratic formula, or completing the square.
In our problem, after using substitution, the equation transformed into a quadratic form: \( y^2 - 3y + 2 = 0 \). This is a straightforward quadratic equation that can often be solved by factoring.

• Factoring involves rewriting the equation as a product of its factors: \((y-2)(y-1) = 0\).
• Each factor can then be solved individually by setting them to zero: \(y-2=0\) and \(y-1=0\).

This results in finding the roots or solutions for \( y \), which are essential to revert back to solving for the original variable \( x \). Knowing how to recognize and factor quadratic equations is a crucial skill in solving various mathematical problems.

Solving Equations

Solving equations involves finding the values of the unknown variables that satisfy the given equation. This can include simple linear equations to more complex non-linear ones.
For our exercise, after returning from a quadratic form to solve for \( y \), we identified \( y = 2 \) and \( y = 1 \). However, only deriving solutions for \( y \) doesn't solve the problem entirely, especially because of the substitution step.

• It's crucial to substitute back to the original terms: \( e^{x} = y \).
• This means translating solutions back to expressions in terms of \( x \): \( x = \ln(2) \) and \( x = \ln(1) \).
• Since \( x = \ln(1) \) is problematic in the expression as it results in a division by zero, it is discarded.

Therefore, you must verify the validity of each solution against the original equation to ensure all angle points of the problem have been considered.

Natural Logarithm

The natural logarithm is a fundamental concept in mathematics, especially in calculus and solving exponential equations. It is the inverse operation of exponentiation when the base is the mathematical constant \( e \) (approximately 2.718).
In the context of our problem, once we've solved \( e^{x} = y \), we use the natural logarithm to determine \( x \). For example:

• If \( e^{x} = 2 \), taking the logarithm results in \( x = \ln(2) \).
• When \( e^{x} = 1 \), since \( \ln(1) = 0 \), solving does not yield a helpful solution.

The natural logarithm provides a way to solve for the exponent in exponential equations, bridging the gap back from our variable substitution to the original equation format.
Understanding the properties of logarithms can aid in maneuvering through seemingly complex equations with confidence.