Problem 25 - Precalculus

Question

Solve the inequality analytically. $$ \frac{1-\ln (x)}{x^{2}}<0 $$

Step-by-Step Solution

Answer

The solution is \( x > e \).

Step 1: Understand the inequality

We need to find the values of \( x \) that satisfy the inequality \( \frac{1 - \ln(x)}{x^2} < 0 \). This means the expression \( \frac{1 - \ln(x)}{x^2} \) must be negative.

Step 2: Identify the sign of x²

Since \( x^2 \) is squared, it is always positive for all \( x \) except at \( x = 0 \). So the denominator \( x^2 > 0 \) (we also exclude zero for division by zero is undefined).

Step 3: Solve the numerator for negativity

The expression \( 1 - \ln(x) \) is negative when \( \ln(x) > 1 \), which is equivalent to \( x > e \) because \( \ln(e) = 1 \).

Step 4: Consider x > 0

Since \( \ln(x) \) and the inequality make sense only for \( x > 0 \) (we already eliminated \( x \leq 0 \) due to division by zero or the logarithm of a non-positive number being undefined), we focus on the range \( x > 0 \).

Step 5: Combine the conditions

The inequality \( \frac{1 - \ln(x)}{x^2} < 0 \) is satisfied if \( 1 - \ln(x) < 0 \) (meaning \( \ln(x) > 1 \)), together with \( x > 0 \). This gives the solution \( x > e \).

Key Concepts

Inequality Solving

Inequality solving is a fundamental skill in precalculus, used to find the values of a variable that satisfy a given condition. In this exercise, we are dealing with the inequality \( \frac{1 - \ln(x)}{x^2} < 0 \). The goal is to determine for which values of \( x \) this inequality holds true.
To solve an inequality analytically, it's crucial to understand what the expression means. The inequality \( \frac{1 - \ln(x)}{x^2} < 0 \) implies that the fraction should be negative. Since \( x^2 \) in the denominator is always positive except at \( x = 0 \) (a point we must avoid as it makes the denominator zero), the sign of the fraction entirely depends on the numerator. The inequality is therefore governed by the condition where \( 1 - \ln(x) < 0 \).
To break it down:

• Ensure the denominator is non-zero and positive.
• Solve the numerator condition \( 1 - \ln(x) < 0 \) by manipulating natural logarithm properties.
• Combine the solved parts to attain the complete solution range, which avoids undefined expressions and satisfies the inequality.

Natural Logarithm

The natural logarithm, commonly represented as \( \ln(x) \), is the inverse function of the exponential function, \( e^x \). It is only defined for positive values of \( x \). Understanding logarithms is crucial in dealing with inequalities like \( \frac{1 - \ln(x)}{x^2} < 0 \). This problem requires us to find when \( 1 - \ln(x) \) becomes negative.
To determine this, consider \( 1 - \ln(x) < 0 \). Rewriting gives \( \ln(x) > 1 \). The expression indicates the point at which the logarithm of \( x \) exceeds 1. Recognizing that \( \ln(e) = 1 \), we note that any \( x > e \) satisfies this inequality. The solution to \( \ln(x) > 1 \) translates directly to \( x > e \), making \( \ln(x) \)'s role pivotal in setting boundaries for \( x \) when solving such problems.
When working with natural logarithms, always consider:

• The domain of \( \ln(x) \): it only operates on positive numbers.
• The properties and common values (e.g., \( \ln(e) = 1 \)).
• How logarithmic inequalities transform into exponential expressions.

Numerator and Denominator Analysis

Numerator and denominator analysis is a strategic method to solve inequalities involving fractions. For \( \frac{1 - \ln(x)}{x^2} < 0 \), solutions hinge on separately assessing the numerator and denominator. This approach simplifies understanding the conditions where an expression is negative, positive, or zero.
In our inequality:

• The denominator \( x^2 \) is always positive when \( x eq 0 \). This is a given due to the squaring, ensuring the sign stays constant in any real case (apart from zero).
• The challenge lies in the numerator \( 1 - \ln(x) \). By solving \( 1 - \ln(x) < 0 \), it implies \( \ln(x) > 1 \), leading to the solution \( x > e \).

By evaluating both parts separately, we determine that for the entire expression to be negative, the numerator alone must be negative since the denominator's positivity is a fixed condition. This is crucial in many inequality solving scenarios, especially when deciphering the behavior of functions across different domains. Always ensure that the denominator never causes division by zero and focus on how the numerator's sign affects the solution set.