Problem 311 - Precalculus

Question

Use logarithms to solve. \(e^{2 x}-e^{x}-132=0\)

Step-by-Step Solution

Answer

The solution is \( x = \ln(12) \).

Step 1: Substitute variable

To simplify the equation, set \( y = e^x \). This substitution means the equation changes to the quadratic form \( y^2 - y - 132 = 0 \).

Step 2: Apply the quadratic formula

Use the quadratic formula to solve for \( y \), which is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -132 \). Substitute these values into the formula:\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-132)}}{2 \times 1} \]Simplify the expression:\[ y = \frac{1 \pm \sqrt{1 + 528}}{2} \]\[ y = \frac{1 \pm \sqrt{529}}{2} \]

Step 3: Find the square root

Calculate the square root of 529, which is 23. Thus, the possible solutions for \( y \) are:\[ y = \frac{1 + 23}{2} = 12 \]\[ y = \frac{1 - 23}{2} = -11 \]

Step 4: Evaluate the feasible solution

Since \( y = e^x \) must be positive, the solution \( y = -11 \) is not feasible. Therefore, \( y = 12 \) is the only feasible solution.

Step 5: Solve for x

With \( y = e^x = 12 \), take the natural logarithm of both sides to find \( x \):\[ x = \ln(12) \]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \( ax^2 + bx + c = 0 \). It allows us to find the values of \( x \) that satisfy the equation. The formula itself is given by:

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

To effectively use the quadratic formula, identifying the coefficients \( a \), \( b \), and \( c \) within the equation is crucial. These coefficients are simply the numbers in front of the terms \( x^2 \), \( x \), and the constant term, respectively.
In our original exercise, we restructured the exponential equation into a quadratic one by substituting variables, specifically setting \( y = e^x \), transforming our equation into \( y^2 - y - 132 = 0 \). This equation is now in the classic quadratic form where:

• \( a = 1 \)
• \( b = -1 \)
• \( c = -132 \)

By plugging these values into the quadratic formula, we solve for \( y \), yielding two solutions, \( y = 12 \) and \( y = -11 \). However, considering the original context, only positive solutions are valid.

Natural Logarithm

Natural logarithms, denoted as \( \ln \), are logarithms with the base of \( e \), where \( e \) is approximately equal to 2.71828. They are very useful in calculus and solving equations involving exponentiation, especially when dealing with growth processes or decay over time.
In our example, once we solved the quadratic equation and found a positive solution for \( y \), specifically \( y = 12 \), we translated this back to terms of \( x \) by recognizing that \( y = e^x \). To solve for \( x \), we took the natural logarithm of both sides:

• \( e^x = 12 \)
• Thus, \( x = \ln(12) \)

Taking the logarithm allows us to effectively "bring down" the exponent, making \( x \) the subject of the equation. Calculating \( \ln(12) \) gives us the exact value of \( x \) that satisfies the initial exponential equation.

Exponential Equations

Exponential equations often involve expressions where variables appear as exponents. Solving these equations frequently requires clever algebraic manipulation or the use of logarithms to expose the variable.
In the given problem, we start with the exponential equation \( e^{2x} - e^x - 132 = 0 \), which presents a challenge because it features exponentials in different powers. We simplify this complexity by substituting \( y = e^x \), thereby linearizing the exponential terms into a more familiar quadratic form, \( y^2 - y - 132 = 0 \).
By using substitutions, logarithms, or even graphing, exponential equations can be transformed and solved methodically -- a handy approach when dealing with problems that at first seem daunting. By following these steps, we can obtain solutions that otherwise might not be immediately obvious, leading to a clearer understanding of how exponential relationships function.