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Ch. 4 - Exponential and Logarithmic Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 5, Problem 95

Solve each equation. 2|ln x|−6=0

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Start with the given equation: \(2|\ln x| - 6 = 0\).
Isolate the absolute value expression by adding 6 to both sides: \(2|\ln x| = 6\).
Divide both sides by 2 to solve for the absolute value: \(|\ln x| = 3\).
Recall that \(|A| = B\) means \(A = B\) or \(A = -B\). So, set up two equations: \(\ln x = 3\) and \(\ln x = -3\).
Solve each equation for \(x\) by exponentiating both sides with base \(e\): \(x = e^{3}\) and \(x = e^{-3}\). Remember to check that \(x > 0\) since the natural logarithm is only defined for positive \(x\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Properties of Logarithmic Functions

The natural logarithm function, denoted ln(x), is defined only for positive values of x. Understanding its domain and behavior is essential when solving equations involving ln(x), as it restricts the possible solutions to x > 0.
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Absolute Value Equations

An absolute value equation like |A| = B splits into two cases: A = B and A = -B. This concept is crucial for solving equations involving absolute values, as it allows us to consider both positive and negative scenarios of the expression inside the absolute value.
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Solving Exponential and Logarithmic Equations

To solve equations involving logarithms, we often isolate the logarithmic expression and then exponentiate both sides to eliminate the log. This process helps convert the equation into a more manageable algebraic form for finding the variable.
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