Problem 26 - Beginning and Intermediate Algebra

Question

Solve the exponential equation algebraically. Then check using a graphing calculator. $$e^{x}-6 e^{-x}=1$$

Step-by-Step Solution

Answer

The short version of the solution to the given exponential equation, \(e^x - 6e^{-x} = 1\), involves using substitution and natural logarithms. Upon solving, we find that \(x \approx \ln{3}\). We can verify this solution using a graphing calculator by graphing the functions \(y_1 = e^x - 1\) and \(y_2 = 6e^{-x}\) and confirming the point of intersection has an x-coordinate equal to the natural logarithm of 3.

Step 1: Rewrite the equation and make it equal to 0

We want to move all terms to one side of the equation to get the following: \(e^x - 6e^{-x} -1 = 0\).

Step 2: Apply substitution to simplify the equation

Make the substitution \(y = e^x\). This will make the equation easier to work with. Then, rewrite the exponential terms in terms of y: \(y - 6\frac{1}{y} - 1 = 0\).

Step 3: Solve for y

Multiply the equation by y to get rid of the fraction: \(y^2 - 6 - y = 0\). Now, change the order: \(y^2 - y - 6 = 0\). Apply the quadratic formula, \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a = 1, b = -1, and c = -6: \(y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} \). Solve for y: \(y \approx -2, 3\).

Step 4: Solve for x

Since \(y = e^x\), we can now solve for x using the natural logarithm function. For y = -2 (We discard this value, as exponential function e^x cannot be negative.) For y = 3: \(x = \ln{3}\).

Step 5: Check the solution using a graphing calculator

Graph the following functions on a graphing calculator: \(y_1 = e^x - 1\) (Note that we moved the 1 to the other side for the graphing calculator) \(y_2 = 6e^{-x}\) Then, find the point of intersection between the two graphs. Verify that the x-coordinate of the point of intersection is approximately equal to the natural logarithm of 3. If so, the solution we found algebraically is correct. The solution to the exponential equation \(e^x - 6e^{-x} = 1\) is \(x \approx \ln{3}\).

Key Concepts

Quadratic Formula

The quadratic formula is a vital tool in algebra for solving quadratic equations, which are equations that can be expressed in the standard form:

• \( ax^2 + bx + c = 0 \)

The quadratic formula allows us to find the value of \(x\), the unknown variable, in such equations and is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our original exercise, after substituting \( y = e^x \) and rewriting the equation as \( y^2 - y - 6 = 0 \), this formula is deployed to solve for \(y\) (which represents \(e^x\)).

• Here, \(a = 1\), \(b = -1\), and \(c = -6\).

Plug these values into the formula to find that \( y \) can be either 3 or -2. However, since \( e^x \) cannot be negative, we discard \( y = -2 \). This leaves us with \( y = 3 \), which we then use to find \( x \). Learning to solve equations using the quadratic formula is a critical step in advancing mathematics skills, providing a strong foundation for tackling more complex problems.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base of Euler's number \(e\), an irrational constant approximately equal to 2.71828. Natural logarithms are incredibly useful when dealing with exponential functions because they serve as the inverse operations of exponentiation.By utilizing the property \( y = e^x \), if we solve for \(x\) when \( y = e^x = 3 \), we take the natural logarithm of both sides to isolate \(x\):

• \(x = \ln{3}\)

The natural logarithm provides a straightforward method to solve for \(x\) when exponential forms are involved. In this problem, \( \ln{3} \) is the value of \(x\) that satisfies the original equation. Understanding natural logarithms is essential for solving a wide array of exponential equations encountered in higher-level mathematics and scientific fields.

Graphing Calculator

A graphing calculator is a powerful tool that allows the visualization of mathematical functions and can be used to verify solutions found algebraically. By graphing the functions in our exercise,

• \( y_1 = e^x - 1 \)
• \( y_2 = 6e^{-x} \)

we can observe where they intersect. The point of intersection confirms the algebraic solution where \( x \approx \ln{3} \). This means the graphs intersect at a point where both functions have the same output, verifying our solution is accurate.

Steps to Use a Graphing Calculator

• Enter the equations into the calculator.
• Graph the two equations.
• Look for points of intersection to verify solutions.

Using a graphing calculator combines visual analysis with algebraic methods, providing a comprehensive approach that ensures understanding and correctness.

Algebraic Solutions

Algebraic solutions involve transforming and manipulating equations using algebraic techniques to find values for unknown variables. The process often includes operations such as substitution, simplification, and factorization.In the exercise, an algebraic approach was used to transform the original exponential equation into a quadratic one. Here’s a recap of key steps:

• Substitution of \( y = e^x \) turns the exponential equation into a more manageable quadratic equation.
• Simplification involves manipulating equations into a standard form that can be solved using known algebraic techniques. In this case, reordering and solving a quadratic equation.
• The Quadratic formula is applied for solutions, selecting only valid values (\( e^x \) cannot be negative).

Algebraic solutions stand as the foundation of solving complex mathematical problems. With practice, these steps become intuitive, enabling solutions to even more challenging equations.