Problem 21 - College Algebra

Question

For the following exercises, 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 the variables

To make the equation easier to solve, let's perform a substitution. Set \( y = e^x \). This turns the original equation \( e^{2x} - e^x - 132 = 0 \) into \( y^2 - y - 132 = 0 \).

Step 2: Solve the quadratic equation

Now we solve the quadratic equation \( y^2 - y - 132 = 0 \). Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -132 \). This gives \( y = \frac{1 \pm \sqrt{1 + 528}}{2} = \frac{1 \pm \sqrt{529}}{2} \).

Step 3: Calculate the roots

Calculate \( \sqrt{529} \), which is 23. Substituting back, we have two potential solutions: \( y = \frac{1 + 23}{2} = 12 \) and \( y = \frac{1 - 23}{2} = -11 \). Since \( y = e^x \) must be positive, we discard \( y = -11 \) as it is not valid for real solutions.

Step 4: Solve for original variable \(x\)

Now solve for \( x \) using \( y = e^x = 12 \). Taking the natural logarithm of both sides gives \( x = \ln{12} \).

Step 5: Verify the solution

Verify by substituting back. Use \( x = \ln{12} \) in the original equation: \( e^{2x} - e^x - 132 = (e^x)^2 - e^x - 132 = 12^2 - 12 - 132 = 144 - 12 - 132 = 0 \) confirms the solution is correct.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and appear frequently in math problems. They take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In many cases, solving quadratic equations requires the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula allows us to find solutions for the variable \( x \) by determining the roots of the equation.

• Roots are the values of \( x \) that satisfy the equation.
• The discriminant, \( b^2 - 4ac \), tells us the nature of the roots. If it's positive, the equation has two real solutions.
• A zero discriminant indicates one real solution, while a negative means the roots are complex.

In this exercise, a substitution \( y = e^x \) transformed the exponential equation into a quadratic one: \( y^2 - y - 132 = 0 \). By applying the quadratic formula, we identified potential roots for \( y \). Only the positive root, \( y = 12 \), is valid because \( y = e^x \), and the exponential function only outputs positive values.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the inverse operation of the exponential function with base \( e \). It is essential in solving equations involving exponentials, like the one in the exercise.

• \( \ln(e^x) = x \) is a key property of natural logarithms.
• The natural logarithm offers a way to "unwrap" exponential expressions, finding the exponent \( x \).

This property is particularly useful when we need to solve for the exponent. For instance, by using \( y = e^x = 12 \), we take the natural logarithm of both sides to solve for \( x \). Doing so gives \( x = \ln{12} \), effectively determining the value of \( x \) that satisfies the original equation.

Exponential Equations

Exponential equations involve variables located in the exponent, such as \( e^{2x} - e^x - 132 = 0 \) from the problem. Solving these can seem challenging without the right approach.

• Identifying exponential equations often involves recognizing the format \( a^{f(x)} = g(x) \), where the variable is in the exponent.
• A common technique is substitution, which can simplify complex expressions. For example, setting \( y = e^x \) transformed the exponential equation into a quadratic, facilitating easier problem resolution.

By converting the problem into a more manageable form, using substitution and logarithms, we can systematically solve exponential equations like the one given. The key is understanding how exponentials behave and utilizing techniques like logarithms to unravel them.