Problem 31 - Precalculus

Question

In Exercises \(1-33,\) solve the equation analytically. $$ e^{x}-3 e^{-x}=2 $$

Step-by-Step Solution

Answer

The solution is \( x = \ln(3) \).

Step 1: Express the Equation in Terms of a Single Exponential Form

Start with the given equation: \[ e^{x} - 3e^{-x} = 2 \] Multiply both sides of the equation by \( e^{x} \) to eliminate the negative exponent, resulting in:\[ e^{2x} - 3 = 2e^{x} \] Rearrange this to form a quadratic in \( e^{x} \):\[ e^{2x} - 2e^{x} - 3 = 0 \]

Step 2: Substitute Variables to Simplify the Equation

Set \( y = e^{x} \). Then the equation becomes:\[ y^2 - 2y - 3 = 0 \] This is a standard quadratic equation, which you can solve using the quadratic formula or by factoring.

Step 3: Factor or Use Quadratic Formula to Solve for y

Attempt factoring the quadratic equation:\[ y^2 - 2y - 3 = (y - 3)(y + 1) = 0 \] The solutions for \( y \) are:\[ y = 3 \quad \text{and} \quad y = -1 \]

Step 4: Back-Substitute to Find x

Recall that \( y = e^{x} \). Therefore, from \( y = 3 \), we have:\[ e^{x} = 3 \]Taking the natural logarithm of both sides gives:\[ x = \ln(3) \]However, \( y = -1 \) is not valid since \( e^{x} \leq 0 \) is not possible for any real \( x \).

Step 5: Solution Verification

Verify the solution by substituting \( x = \ln(3) \) back into the original equation.Substituting \( x = \ln(3) \) into the equation gives:\[ e^{\ln(3)} - 3e^{-\ln(3)} = 3 - \frac{3}{3} = 3 - 1 = 2 \]The solution satisfies the original equation.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, which means its highest exponent is 2. The general form is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula. In our exercise, we transformed the original exponential equation into a quadratic form by setting \( y = e^x \). This resulted in the equation \( y^2 - 2y - 3 = 0 \).

Factoring this equation, we found the solutions \( y - 3 = 0 \) and \( y + 1 = 0 \), which gave us \( y = 3 \) and \( y = -1 \). Solving quadratic equations this way often gives us two solutions, but they must be checked for validity in the context of the problem. Here, only \( y = 3 \) was valid, since \( y = -1 \) would imply a negative value for an exponential function, which isn't possible with real numbers.

Key points when dealing with quadratics include:

• The solutions are the values of the variable that make the equation equal to zero.
• The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is a powerful tool to find solutions when factoring is complex.
• Ensure that the solutions make sense in the context of your problem.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately 2.718. It is a special logarithm that appears frequently in calculus as it simplifies many equations involving exponential functions. In the solution to our problem, we used the natural logarithm to solve for \( x \) from \( y = e^x \).

We applied \( \ln \) to both sides of the equation \( e^x = 3 \), resulting in \( x = \ln(3) \). Using the natural logarithm here helped us solve the equation analytically, providing a straightforward path to find \( x \).

Important concepts when using natural logarithms include:

• \( \ln(e) = 1 \) because \( e^1 = e \).
• \( \ln(1) = 0 \) since \( e^0 = 1 \).
• The logarithm allows us to solve exponential equations where the unknown is an exponent.

Exponential Function

Exponential functions are vital in many fields such as mathematics, biology, and finance, as they describe growth and decay processes. The general form of an exponential function is \( f(x) = a \cdot e^{kx} \), where \( a \) and \( k \) are constants, and \( e \) is the base of the natural logarithms.

In the problem given, we started with the equation \( e^x - 3e^{-x} = 2 \). This equation involves exponential functions that have both positive and negative exponents. Such expressions can be simplified by multiplying through by \( e^x \), which allows us to eliminate negative exponents.

Once transformed, the equation took the quadratic form that was easier to solve. Exponential functions have unique properties:

• They are always positive for real exponents, i.e., \( e^x > 0 \) for all real \( x \).
• Given their continuous and smooth nature, they can model both exponential growth \((k > 0)\) and decay \((k < 0)\).
• They serve as the inverse of the natural logarithm, allowing for flexible function transformations.

Understanding these properties is crucial when solving exponential equations analytically.