Problem 95 - College Algebra

Question

Solve each equation. $$ 2|\ln x|-6=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = e^3\) and \(x = e^{-3}\)

Step 1: Isolate the Absolute Value

Rewrite the equation to isolate the absolute value on one side of the equation. This would be \(2|\ln x|=6\).

Step 2: Simplify the Equation

Let's divide both sides by 2 to further simplify the equation. This leads us to \(|\ln x|=3\).

Step 3: Handle the Absolute Value

Since we have an absolute value, we will have two possible solutions. This gives us two equations: \(\ln x = 3\) and \(\ln x = -3\).

Step 4: Solve for \(x\)

Applying the property of logarithms to the first equation, we get \(x = e^3\). Applying the same property to the second equation, we get \(x = e^{-3}\).

Key Concepts

Understanding Absolute Value in Equations

When solving equations involving absolute value, the first step is to isolate the absolute value itself. Absolute value represents the distance a number is from zero on the number line, regardless of direction. In this context, \( |\ln x| \) is the absolute value of the logarithm of \( x \). Since the absolute value can represent both a positive and a negative scenario, you'll have two possible equations to solve. These are derived from setting \( \ln x \) equal to both the positive and negative values of the isolated number. In our equation, \( 2|\ln x|-6=0 \), isolating leads to \( |\ln x|=3 \). Thus, the two scenarios are \( \ln x = 3 \) and \( \ln x = -3 \). This provides a broader set of potential solutions.

Exploring Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. The equation \( \ln x \) signifies the natural logarithm of \( x \), which is the power to which \( e \) (approximately 2.718) must be raised to produce \( x \). In simpler terms, if \( \ln x = 3 \), it means \( e^3 = x \). Logarithmic equations are crucial for solving for \( x \) in situations where \( x \) is an exponent itself. They transform multiplicative relationships into additive ones, simplifying the process of finding unknowns in equations. By addressing \( \ln x = 3 \) and \( \ln x = -3 \), we efficiently find two solutions using exponential equivalences: \( x = e^3 \) and \( x = e^{-3} \). This illustrates the transformative power of logarithms in mathematics.

Applying Properties of Logarithms

The properties of logarithms make solving equations with logarithmic expressions manageable and systematic. Key properties to remember include:

• The logarithm of a power: \( \log_b(a^n) = n \log_b(a) \)
• The exponential log equivalence: \( b^{\log_b(y)} = y \)

These properties allow you to switch between logarithmic and exponential forms seamlessly. In our exercise, once \( |\ln x| = 3 \) is set, translating \( \ln x = 3 \) into \( x = e^3 \) utilizes the equivalence between logarithmic and exponential operations. Recognizing these properties enables a student to break down complex expressions and equations, paving a straightforward path to finding solutions. By using these properties carefully, equations that initially seem daunting break down into smaller, more manageable parts.