Problem 30 - Precalculus

Question

Solve the inequality analytically. $$ \ln \left(x^{2}\right) \leq(\ln (x))^{2} $$

Step-by-Step Solution

Answer

The solution to the inequality is \(x \in [1, e^2] \cup (e^2, \infty)\).

Step 1: Rewrite the Given Inequality

The equation is given as \( \ln(x^2) \leq (\ln(x))^2 \). By using the power rule of logarithms, simplify the left-hand side: \( \ln(x^2) = 2 \ln(x) \). Thus, the inequality becomes: \[2 \ln(x) \leq (\ln(x))^2\].

Step 2: Rearrange the Inequality

Subtract \(2 \ln(x)\) from both sides of the inequality, which gives:\[0 \leq (\ln(x))^2 - 2 \ln(x)\].This rewritten form helps us see the inequality as a quadratic expression in \(\ln(x)\).

Step 3: Set Up Quadratic Inequality

Consider it as:\[(\ln(x))^2 - 2 \ln(x) \geq 0\].Let \(y = \ln(x)\). Substitute into the inequality, which gives:\[y^2 - 2y \geq 0\].

Step 4: Factor the Quadratic

Factor the quadratic expression:\[(y - 0)(y - 2) \geq 0\].The solution involves finding when the product of these factors is non-negative.

Step 5: Determine Critical Points and Sign Analysis

Set each factor to zero to find critical points:- \(y - 0 = 0 \Rightarrow y = 0\)- \(y - 2 = 0 \Rightarrow y = 2\).Create intervals based on these points for sign analysis:1. \((-\infty, 0)\)2. \((0, 2)\)3. \((2, \infty)\).Perform a sign test in each interval to determine where the inequality holds.

Step 6: Test Intervals

Choose test values from each interval to determine signs:- Interval \((-\infty, 0)\): Choose \(y = -1\). \((-1-0)(-1-2) > 0\) is false.- Interval \((0, 2)\): Choose \(y = 1\). \((1-0)(1-2) < 0\) is false.- Interval \((2, \infty)\): Choose \(y = 3\). \((3-0)(3-2) > 0\) is true.Equality holds at \(y = 0\) and \(y = 2\).

Step 7: Translate Back to x

Convert the solutions for \(y\) back to \(x\):- For \(y = 0\), \(\ln(x) = 0\) implies \(x = e^0 = 1\).- For \(y = 2\), \(\ln(x) = 2\) implies \(x = e^2\).Thus, the inequality holds for \(x \in [1, e^2] \cup (e^2, \infty)\).

Key Concepts

Logarithms

When solving problems involving logarithms, it's important to understand what a logarithm is. A logarithm is the inverse operation of exponentiation. Simply put, if you have a number in the form of a base raised to an exponent that equals another number, the logarithm helps you find the exponent.
For example, consider the base 10 logarithm of 100, which is 2, because 10 raised to the power of 2 equals 100. Similarly, the natural logarithm, denoted by \(\ln\), uses the base \(e\) (approximately 2.718).
Understanding this allows us to approach logarithmic inequalities logically, as it helps us see their potential transformations, simplify them, and establish ranges for solutions.

Quadratic Inequalities

Quadratic inequalities are similar to quadratic equations but they involve inequalities (like \(<, >, \leq, \geq\)) instead of equalities. The general method for solving them is to rearrange the equation into a standard quadratic form and then analyze it.
For example, consider the inequality \[(\ln(x))^2 - 2 \ln(x) \geq 0\]. We recognize this as 'quadratic' because it has a squared term, a linear term, and a constant term when rearranged.
Factoring is a powerful tool with quadratics, often turning complex expressions into products of simpler binomials, which can then be further analyzed through sign tests to find solution ranges.

Sign Analysis

Sign analysis is a technique used to determine where an inequality holds true. Once a quadratic inequality is factored, such as \[(y - 0)(y - 2) \geq 0\], we identify the critical points, \(y = 0\) and \(y = 2\). These points divide the number line into intervals.

• The first interval is \(( -\infty, 0)\).
• The second is \((0, 2)\).
• The third is \((2, \infty)\).

To evaluate each interval, pick any number within it as a test point. Substitute it back into the factored form of the inequality to determine if the product of the terms is positive or negative.
This approach effectively pinpoints the exact intervals where the inequality holds true or is satisfied.

Power Rule of Logarithms

The power rule of logarithms is a useful property that simplifies the manipulation of logarithmic expressions. It states that \(\ln(x^a) = a \cdot \ln(x)\). Through this rule, we can transform logarithms of powers into products, which is particularly helpful in solving logarithmic inequalities.
In our problem, initially given as \(\ln(x^2) \leq (\ln(x))^2\), we applied the power rule to simplify \(\ln(x^2)\) to \(2 \cdot \ln(x)\), making the subsequent steps in solving the inequality more straightforward.
Knowing and applying these logarithmic properties fluently allows for a smoother transition from complex expressions to simpler forms, aiding in finding solutions effectively.