Logarithmic Inequalities: Theory, Solution Methods and Practice Problems

A logarithmic inequality is an inequality in which the unknown appears as the argument of at least one logarithm. Solving such inequalities requires, beyond proficiency in algebraic manipulation, a rigorous treatment of two distinct elements: the existence conditions, which define the domain of the inequality, and the direction of the inequality, which depends crucially on the monotonicity of the logarithmic function with respect to its base. An error in the direction — as common as it is serious — yields incorrect solution sets even when every algebraic step is impeccable.

Table of Contents

• Review of the Logarithmic Function and Monotonicity
• Domain of a Logarithmic Inequality
• Elementary Logarithmic Inequalities
• Inequalities Involving Sums or Differences of Logarithms
• Inequalities Comparing Two Logarithms
• Inequalities with Logarithms of Different Bases
• Inequalities Solvable by Substitution
• General Solution Procedure
• Practice Problems
• Graphical Interpretation

The solution of logarithmic inequalities depends directly and unavoidably on the monotonicity of the logarithmic function. It is therefore essential to recall this property precisely before proceeding.

Given a base \( a \in \mathbb{R} \) with \( a > 0 \) and \( a \neq 1 \), the logarithmic function \[ \log_a \colon (0, +\infty) \longrightarrow \mathbb{R}, \qquad x \longmapsto \log_a x \] is strictly monotone. More precisely:

• it is strictly increasing when \( a > 1 \): for every \( u, v \in (0, +\infty) \), \[ u < v \quad \Longleftrightarrow \quad \log_a u < \log_a v; \]
• it is strictly decreasing when \( 0 < a < 1 \): for every \( u, v \in (0, +\infty) \), \[ u < v \quad \Longleftrightarrow \quad \log_a u > \log_a v. \]

This dichotomy is the logical foundation of the entire theory of logarithmic inequalities. When the base exceeds one, applying the logarithmic function to both sides of an inequality preserves its direction; when the base lies between zero and one, it reverses it. Overlooking this step — reversing the direction when it should be reversed, or failing to do so when required — is the most frequent conceptual error and leads to systematically incorrect solution sets.

We also recall that the natural domain of the logarithm is \( (0, +\infty) \): the real logarithm of a non-positive number is undefined. This restriction is the source of all existence conditions in logarithmic inequalities, and its treatment is the subject of the following section.

The domain of a logarithmic inequality is the set of real values of the unknown for which every expression appearing in the inequality is well defined. For each term \( \log_a f_i(x) \), as \( i \) ranges over the index set \( I \) of all logarithms present in the inequality, one must impose: \[ f_i(x) > 0. \]

The domain of the inequality is the intersection of all these conditions: \[ \mathcal{D} = \bigcap_{i \in I} \bigl\{ x \in \mathbb{R} \mid f_i(x) > 0 \bigr\}. \]

Fundamental rule. The solution set of the inequality is a subset of \( \mathcal{D} \). Any value of the unknown that formally satisfies the algebraic inequality obtained through transformations, but does not belong to \( \mathcal{D} \), must be discarded.

Unlike the situation for equations — where one checks whether individual values belong to \( \mathcal{D} \) — in inequalities one must intersect the set of algebraic solutions (typically an interval or a union of intervals) with \( \mathcal{D} \). This intersection constitutes the set of admissible solutions.

It is methodologically essential to determine \( \mathcal{D} \) before any algebraic manipulation, so as to keep in mind throughout the set within which solutions must be sought.

A logarithmic inequality is called elementary if it can be reduced to the standard form: \[ \log_a f(x) \;\square\; k, \qquad k \in \mathbb{R}, \] where \( \square \) denotes one of the symbols \( >, \geq, <, \leq \). The solution method consists in inverting the logarithmic function, taking into account its monotonicity with respect to the base \( a \).

Case \( a > 1 \) (increasing function). Since \( \log_a \) is strictly increasing, the direction of the inequality is preserved under inversion:

• \( \log_a f(x) > k \;\Longleftrightarrow\; f(x) > a^k \). Since \( a^k > 0 \), the condition \( f(x) > a^k \) automatically implies \( f(x) > 0 \), so the solutions of \( f(x) > a^k \) already lie within the domain.
• \( \log_a f(x) < k \;\Longleftrightarrow\; f(x) < a^k \). Since \( f(x) < a^k \) does not imply \( f(x) > 0 \), one must intersect with the domain, yielding the equivalent condition \( 0 < f(x) < a^k \).

Case \( 0 < a < 1 \) (decreasing function). Since \( \log_a \) is strictly decreasing, the direction of the inequality is reversed under inversion:

• \( \log_a f(x) > k \;\Longleftrightarrow\; f(x) < a^k \). Since \( f(x) < a^k \) does not imply \( f(x) > 0 \), one must intersect with the domain, yielding \( 0 < f(x) < a^k \).
• \( \log_a f(x) < k \;\Longleftrightarrow\; f(x) > a^k \). Since \( a^k > 0 \), the condition \( f(x) > a^k \) automatically implies \( f(x) > 0 \), so the solutions already lie within the domain.

Note the dual structure: the domain condition is automatically satisfied whenever the direction of the inequality (after inversion) is \( f(x) > a^k \), whereas it requires an explicit intersection when the direction is \( f(x) < a^k \). In the latter case the complete condition is \( 0 < f(x) < a^k \), which expresses the conjunction of the domain requirement with the algebraic inequality. The same scheme applies to \( \geq \) and \( \leq \), the only difference being that strict inequalities become non-strict.

Example 1. Solve \( \log_3(2x - 1) > 2 \).

Domain. \( 2x - 1 > 0 \Rightarrow \mathcal{D} = \bigl(\tfrac{1}{2}, +\infty\bigr) \).

Solution. The base is \( a = 3 > 1 \): the function is increasing, so the direction is preserved. \[ 2x - 1 > 3^2 = 9 \quad \Longrightarrow \quad 2x > 10 \quad \Longrightarrow \quad x > 5. \] Since \( 5 > \tfrac{1}{2} \), the solutions \( x > 5 \) are already contained in \( \mathcal{D} \).

Solution set: \( (5, +\infty) \).

Example 2. Solve \( \log_3(2x - 1) < 2 \).

Domain. \( \mathcal{D} = \bigl(\tfrac{1}{2}, +\infty\bigr) \).

Solution. Base \( a = 3 > 1 \), increasing function, direction preserved: \[ 2x - 1 < 9 \quad \Longrightarrow \quad x < 5. \] Since \( 2x - 1 < 9 \) does not guarantee \( 2x - 1 > 0 \), we intersect with \( \mathcal{D} \): \[ \frac{1}{2} < x < 5. \]

Solution set: \( \bigl(\tfrac{1}{2}, 5\bigr) \).

Example 3. Solve \( \log_{1/3}(x + 1) > 1 \).

Domain. \( x + 1 > 0 \Rightarrow \mathcal{D} = (-1, +\infty) \).

Solution. The base is \( a = \tfrac{1}{3} \), with \( 0 < a < 1 \): the function is decreasing, so the direction is reversed. \[ x + 1 < \left(\frac{1}{3}\right)^1 = \frac{1}{3}. \] Since \( x + 1 < \tfrac{1}{3} \) does not imply \( x + 1 > 0 \), we intersect with the domain: \[ 0 < x + 1 < \frac{1}{3} \quad \Longrightarrow \quad -1 < x < -\frac{2}{3}. \]

Solution set: \( \bigl(-1, -\tfrac{2}{3}\bigr) \).

Example 4. Solve \( \log_{1/2}(x - 3) < -2 \).

Domain. \( x - 3 > 0 \Rightarrow \mathcal{D} = (3, +\infty) \).

Solution. Base \( a = \tfrac{1}{2} \), with \( 0 < a < 1 \): decreasing function, direction reversed: \[ x - 3 > \left(\frac{1}{2}\right)^{-2} = 4 \quad \Longrightarrow \quad x > 7. \] Since \( x > 7 \) implies \( x - 3 > 0 \), the solutions already lie within \( \mathcal{D} \).

Solution set: \( (7, +\infty) \).

When an inequality contains a sum or difference of logarithms with the same base, the product and quotient rules are applied to consolidate them into a single logarithm, thereby reducing the inequality to elementary form. As already noted, the domain conditions must be imposed on the original arguments, not on the argument of the resulting merged logarithm.

Warning. The identity \( \log_a f(x) + \log_a g(x) = \log_a[f(x)g(x)] \) is valid only when \( f(x) > 0 \) and \( g(x) > 0 \) separately. The product \( f(x)g(x) \) can be positive even when both factors are negative; in that case the logarithm of the product would be defined in the transformed inequality, while the logarithms of the individual factors would not be defined in the original. This is why the domain must be determined from the original arguments.

Example 1. Solve \( \log_2 x + \log_2(x - 2) > 3 \).

Domain. \( x > 0 \) and \( x - 2 > 0 \), so \( \mathcal{D} = (2, +\infty) \).

Solution. Applying the product rule: \[ \log_2[x(x-2)] > 3. \] Base \( a = 2 > 1 \): direction preserved. \[ x(x - 2) > 2^3 = 8 \quad \Longrightarrow \quad x^2 - 2x - 8 > 0 \quad \Longrightarrow \quad (x-4)(x+2) > 0. \] The quadratic is positive for \( x < -2 \) or \( x > 4 \). Intersecting with \( \mathcal{D} = (2, +\infty) \): \[ x > 4. \]

Solution set: \( (4, +\infty) \).

Example 2. Solve \( \log_3(x + 5) - \log_3(x - 1) > 1 \).

Domain. \( x + 5 > 0 \) and \( x - 1 > 0 \), so \( \mathcal{D} = (1, +\infty) \).

Solution. Applying the quotient rule: \[ \log_3\!\left(\frac{x+5}{x-1}\right) > 1. \] Base \( a = 3 > 1 \): direction preserved. \[ \frac{x + 5}{x - 1} > 3. \] Since we are in the domain \( x > 1 \), we have \( x - 1 > 0 \), so both sides may be multiplied by \( x - 1 \) without reversing the direction: \[ x + 5 > 3(x - 1) \quad \Longrightarrow \quad x + 5 > 3x - 3 \quad \Longrightarrow \quad 8 > 2x \quad \Longrightarrow \quad x < 4. \] Intersecting with \( \mathcal{D} = (1, +\infty) \): \[ 1 < x < 4. \]

Solution set: \( (1, 4) \).

Methodological note. In Example 2, multiplication by \( x - 1 \) is valid — without reversing the direction — precisely because we are operating within the domain, where \( x - 1 > 0 \) is guaranteed. Outside the domain this step would require consideration of the sign of the denominator, considerably complicating the procedure. This provides a further argument in favour of determining \( \mathcal{D} \) at the outset.

When the inequality takes the form: \[ \log_a f(x) \;\square\; \log_a g(x), \] the strict monotonicity of the function \( t \mapsto \log_a t \) on \( (0, +\infty) \) allows the logarithm to be eliminated, subject to the direction dictated by the base:

• If \( a > 1 \) (increasing function): \( \log_a f(x) > \log_a g(x) \Longleftrightarrow f(x) > g(x) \), provided both arguments are positive.
• If \( 0 < a < 1 \) (decreasing function): \( \log_a f(x) > \log_a g(x) \Longleftrightarrow f(x) < g(x) \), subject to the same domain conditions.

In both cases the solution procedure is as follows: determine \( \mathcal{D} \) by imposing \( f(x) > 0 \) and \( g(x) > 0 \), eliminate the logarithm by applying the appropriate correspondence (with or without reversal of direction), solve the resulting algebraic inequalities, and finally intersect the solution set with \( \mathcal{D} \).

Logical remark. Within the domain \( \mathcal{D} \), both conditions \( f(x) > 0 \) and \( g(x) > 0 \) are guaranteed by assumption. Therefore, when eliminating the logarithm, no additional positivity conditions need to be imposed: they are already built into \( \mathcal{D} \). The final intersection with \( \mathcal{D} \) is thus sufficient to ensure the correctness of the solutions.

Example 1. Solve \( \log_2(x + 3) > \log_2(2x - 1) \).

Domain. \( x + 3 > 0 \) and \( 2x - 1 > 0 \), so \( \mathcal{D} = \bigl(\tfrac{1}{2}, +\infty\bigr) \).

Solution. Base \( a = 2 > 1 \): direction preserved. \[ x + 3 > 2x - 1 \quad \Longrightarrow \quad 4 > x \quad \Longrightarrow \quad x < 4. \] Intersecting with \( \mathcal{D} = \bigl(\tfrac{1}{2}, +\infty\bigr) \): \[ \frac{1}{2} < x < 4. \]

Solution set: \( \bigl(\tfrac{1}{2}, 4\bigr) \).

Example 2. Solve \( \log_{1/2}(3 - x) > \log_{1/2}(x + 1) \).

Domain. \( 3 - x > 0 \) and \( x + 1 > 0 \), so \( \mathcal{D} = (-1, 3) \).

Solution. Base \( a = \tfrac{1}{2} \), with \( 0 < a < 1 \): decreasing function, direction reversed. \[ 3 - x < x + 1 \quad \Longrightarrow \quad 2 < 2x \quad \Longrightarrow \quad x > 1. \] Intersecting with \( \mathcal{D} = (-1, 3) \): \[ 1 < x < 3. \]

Solution set: \( (1, 3) \).

Example 3. Solve \( \log_5(x^2 - 3x) \geq \log_5(x + 7) \).

Domain. \( x^2 - 3x > 0 \) and \( x + 7 > 0 \). Factoring: \( x(x - 3) > 0 \) when \( x < 0 \) or \( x > 3 \). The second condition gives \( x > -7 \). Hence \( \mathcal{D} = (-7, 0) \cup (3, +\infty) \).

Solution. Base \( a = 5 > 1 \): direction preserved (with \( \geq \)). \[ x^2 - 3x \geq x + 7 \quad \Longrightarrow \quad x^2 - 4x - 7 \geq 0. \] The roots of the quadratic are \( x = 2 \pm \sqrt{11} \). The quadratic is non-negative for \( x \leq 2 - \sqrt{11} \) or \( x \geq 2 + \sqrt{11} \). Since \( 2 - \sqrt{11} \approx -1{.}32 \) and \( 2 + \sqrt{11} \approx 5{.}32 \), intersecting with \( \mathcal{D} = (-7, 0) \cup (3, +\infty) \): \[ \bigl(-7, 2 - \sqrt{11}\,\bigr] \cup \bigl[2 + \sqrt{11}, +\infty\bigr). \]

Solution set: \( \bigl(-7, 2 - \sqrt{11}\,\bigr] \cup \bigl[2 + \sqrt{11}, +\infty\bigr) \).

When an inequality contains logarithms with different bases, neither the comparison principle nor the operational properties can be applied directly. The standard method is to convert all logarithms to a common base using the change-of-base formula: \[ \log_a x = \frac{\log_b x}{\log_b a}, \] where the auxiliary base \( b \) is chosen to simplify the calculations. The most common choices are \( b = 10 \) (common logarithm) or \( b = e \) (natural logarithm); in many cases it is convenient to take as the common base one already present in the inequality.

Warning. When both sides of an inequality are multiplied or divided by a quantity, the sign of that quantity must be taken into account. In particular, \( \log_b a \) has a definite sign: it is positive when \( b \) and \( a \) are both greater than one or both less than one (i.e., on the same side of 1), and negative otherwise. Multiplying by a negative quantity reverses the direction of the inequality.

Example 1. Solve \( \log_2 x > \log_4(x + 2) \).

Domain. \( x > 0 \) and \( x + 2 > 0 \), so \( \mathcal{D} = (0, +\infty) \).

Solution. We convert \( \log_4(x+2) \) to base 2: \[ \log_4(x + 2) = \frac{\log_2(x + 2)}{\log_2 4} = \frac{\log_2(x + 2)}{2}. \] The inequality becomes: \[ \log_2 x > \frac{\log_2(x+2)}{2}. \] Multiplying both sides by 2 (positive, direction preserved): \[ 2\log_2 x > \log_2(x + 2) \quad \Longrightarrow \quad \log_2 x^2 > \log_2(x + 2). \] The identity \( 2\log_2 x = \log_2 x^2 \) is valid since \( x > 0 \) in the domain. Since the base 2 exceeds one, the direction is preserved: \[ x^2 > x + 2 \quad \Longrightarrow \quad x^2 - x - 2 > 0 \quad \Longrightarrow \quad (x - 2)(x + 1) > 0. \] The quadratic is positive for \( x < -1 \) or \( x > 2 \). Intersecting with \( \mathcal{D} = (0, +\infty) \): \[ x > 2. \]

Solution set: \( (2, +\infty) \).

Example 2. Solve \( \log_2 x + \log_4 x \leq 3 \).

Domain. \( \mathcal{D} = (0, +\infty) \).

Solution. We convert \( \log_4 x \) to base 2: \[ \log_4 x = \frac{\log_2 x}{2}. \] Setting \( t = \log_2 x \): \[ t + \frac{t}{2} \leq 3 \quad \Longrightarrow \quad \frac{3t}{2} \leq 3 \quad \Longrightarrow \quad t \leq 2. \] Returning to the original variable: \( \log_2 x \leq 2 \Rightarrow x \leq 4 \). Intersecting with \( \mathcal{D} = (0, +\infty) \): \[ 0 < x \leq 4. \]

Solution set: \( (0, 4] \).

An important class of logarithmic inequalities consists of those in which the logarithm appears as the argument of a polynomial expression. The typical form is: \[ P\!\bigl(\log_a f(x)\bigr) \;\square\; 0, \] where \( P \) is a polynomial. The method consists in setting \( t = \log_a f(x) \), solving the algebraic inequality \( P(t) \;\square\; 0 \) in \( t \) to determine the admissible range of values for \( t \), and then, for each resulting interval, solving the corresponding elementary inequality \( \log_a f(x) \;\square\; t_k \).

Note. The substitution \( t = \log_a f(x) \) transfers the inequality to the auxiliary variable \( t \). Once the inequality in \( t \) has been solved, each resulting interval must be translated back into a corresponding condition on \( x \). Thus, an interval \( t \in [t_1, t_2] \) becomes the conjunction of the two elementary inequalities \( \log_a f(x) \geq t_1 \) and \( \log_a f(x) \leq t_2 \). When the solution in \( t \) is a union of intervals, each interval must be handled separately and the resulting solution sets combined.

Example 1. Solve \( (\log_3 x)^2 - \log_3 x - 2 \geq 0 \).

Domain. \( x > 0 \), so \( \mathcal{D} = (0, +\infty) \).

Substitution. Let \( t = \log_3 x \). The inequality becomes: \[ t^2 - t - 2 \geq 0 \quad \Longrightarrow \quad (t - 2)(t + 1) \geq 0. \] The quadratic is non-negative for \( t \leq -1 \) or \( t \geq 2 \).

Returning to the original variable.

• \( \log_3 x \leq -1 \): base \( 3 > 1 \), increasing function, direction preserved: \( x \leq 3^{-1} = \tfrac{1}{3} \). Intersecting with \( \mathcal{D} \): \( 0 < x \leq \tfrac{1}{3} \).
• \( \log_3 x \geq 2 \): \( x \geq 3^2 = 9 \). Already in \( \mathcal{D} \).

Solution set: \( \bigl(0, \tfrac{1}{3}\bigr] \cup [9, +\infty) \).

Example 2. Solve \( (\log_2 x)^2 - 4 < 0 \).

Domain. \( \mathcal{D} = (0, +\infty) \).

Substitution. Let \( t = \log_2 x \): \[ t^2 - 4 < 0 \quad \Longrightarrow \quad (t-2)(t+2) < 0 \quad \Longrightarrow \quad -2 < t < 2. \]

Returning to the original variable. The condition \( -2 < \log_2 x < 2 \) decomposes as: \[ \log_2 x > -2 \;\Rightarrow\; x > 2^{-2} = \tfrac{1}{4}, \qquad \log_2 x < 2 \;\Rightarrow\; x < 4. \] Intersecting with \( \mathcal{D} = (0, +\infty) \): \[ \frac{1}{4} < x < 4. \]

Solution set: \( \bigl(\tfrac{1}{4}, 4\bigr) \).

Example 3. Solve \( 2(\log_5 x)^2 + 3\log_5 x - 2 < 0 \).

Domain. \( \mathcal{D} = (0, +\infty) \).

Substitution. Let \( t = \log_5 x \): \[ 2t^2 + 3t - 2 < 0 \quad \Longrightarrow \quad t = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}. \] The roots are \( t_1 = \tfrac{1}{2} \) and \( t_2 = -2 \). The quadratic is negative for \( -2 < t < \tfrac{1}{2} \).

Returning to the original variable. \[ -2 < \log_5 x < \frac{1}{2} \quad \Longrightarrow \quad 5^{-2} < x < 5^{1/2} \quad \Longrightarrow \quad \frac{1}{25} < x < \sqrt{5}. \] Already in \( \mathcal{D} \).

Solution set: \( \Bigl(\dfrac{1}{25}, \sqrt{5}\Bigr) \).

The following procedure constitutes a complete and rigorous protocol applicable to any type of logarithmic inequality covered in this work.

1. Determine the domain. For each logarithm \( \log_a f_i(x) \) present in the inequality, impose \( f_i(x) > 0 \). Compute \( \mathcal{D} = \bigcap_{i \in I} \{x \in \mathbb{R} \mid f_i(x) > 0\} \).
2. Convert to a common base (if necessary). If logarithms with different bases are present, apply the change-of-base formula to unify them, paying attention to the sign of \( \log_b a \) in subsequent operations on the inequality.
3. Simplify using logarithm properties. Apply the product, quotient, and power rules — valid only for positive arguments — to reduce the inequality to one of the standard forms: \( \log_a f(x) \,\square\, k \), or \( \log_a f(x) \,\square\, \log_a g(x) \), or \( P(\log_a f(x)) \,\square\, 0 \).
4. Eliminate the logarithm. Apply the monotonicity correspondence: if \( a > 1 \), the direction is preserved; if \( 0 < a < 1 \), it is reversed. For polynomial forms, set \( t = \log_a f(x) \) and solve the resulting algebraic inequality in \( t \).
5. Solve the resulting algebraic inequality. Determine the solution set (typically an interval or a union of intervals).
6. Intersect with the domain. The solution set of the logarithmic inequality is the intersection of the algebraic solution set with \( \mathcal{D} \). Verify that the positivity conditions on the arguments are satisfied; discard any portion lying outside \( \mathcal{D} \).
7. State the solution set.

Exercise 1. Solve \( \log_2(x + 3) \geq 4 \).

Domain. \( x + 3 > 0 \Rightarrow \mathcal{D} = (-3, +\infty) \).

Solution. Base \( 2 > 1 \): direction preserved. \( x + 3 \geq 2^4 = 16 \Rightarrow x \geq 13 \). Already in \( \mathcal{D} \).

Solution set: \( [13, +\infty) \).

Exercise 2. Solve \( \log_3 x + \log_3(x - 1) < 1 \).

Domain. \( x > 0 \) and \( x - 1 > 0 \Rightarrow \mathcal{D} = (1, +\infty) \).

Solution. \[ \log_3[x(x-1)] < 1 \quad \Longrightarrow \quad x(x-1) < 3 \quad \Longrightarrow \quad x^2 - x - 3 < 0. \] The roots are \( x = \dfrac{1 \pm \sqrt{13}}{2} \). The quadratic is negative for \( \dfrac{1 - \sqrt{13}}{2} < x < \dfrac{1 + \sqrt{13}}{2} \). Intersecting with \( \mathcal{D} = (1, +\infty) \): \[ 1 < x < \frac{1 + \sqrt{13}}{2}. \]

Solution set: \( \Bigl(1,\, \dfrac{1 + \sqrt{13}}{2}\Bigr) \).

Exercise 3. Solve \( \log_5(x + 1) \geq \log_5(2x - 3) \).

Domain. \( x + 1 > 0 \) and \( 2x - 3 > 0 \Rightarrow \mathcal{D} = \bigl(\tfrac{3}{2}, +\infty\bigr) \).

Solution. Base \( 5 > 1 \): direction preserved. \[ x + 1 \geq 2x - 3 \quad \Longrightarrow \quad 4 \geq x \quad \Longrightarrow \quad x \leq 4. \] Intersecting with \( \mathcal{D} = \bigl(\tfrac{3}{2}, +\infty\bigr) \): \[ \frac{3}{2} < x \leq 4. \]

Solution set: \( \bigl(\tfrac{3}{2},\, 4\bigr] \).

Exercise 4. Solve \( \log_2(x - 1) + \log_2(x - 5) < 3 \).

Domain. \( x - 1 > 0 \) and \( x - 5 > 0 \Rightarrow \mathcal{D} = (5, +\infty) \).

Solution. \[ \log_2[(x-1)(x-5)] < 3 \quad \Longrightarrow \quad (x-1)(x-5) < 8. \] \[ x^2 - 6x + 5 < 8 \quad \Longrightarrow \quad x^2 - 6x - 3 < 0. \] The roots are \( x = 3 \pm 2\sqrt{3} \). The quadratic is negative for \( 3 - 2\sqrt{3} < x < 3 + 2\sqrt{3} \). Since \( 3 + 2\sqrt{3} \approx 6{.}46 \), intersecting with \( \mathcal{D} = (5, +\infty) \): \[ 5 < x < 3 + 2\sqrt{3}. \]

Solution set: \( \bigl(5,\, 3 + 2\sqrt{3}\bigr) \).

Exercise 5. Solve \( \log_{1/2}(x + 1) < \log_{1/2}(3 - x) \).

Domain. \( x + 1 > 0 \) and \( 3 - x > 0 \Rightarrow \mathcal{D} = (-1, 3) \).

Solution. Base \( \tfrac{1}{2} \), with \( 0 < a < 1 \): decreasing function, direction reversed. \[ x + 1 > 3 - x \quad \Longrightarrow \quad 2x > 2 \quad \Longrightarrow \quad x > 1. \] Intersecting with \( \mathcal{D} = (-1, 3) \): \[ 1 < x < 3. \]

Solution set: \( (1, 3) \).

Exercise 6. Solve \( (\log_2 x)^2 - 5\log_2 x + 6 < 0 \).

Domain. \( \mathcal{D} = (0, +\infty) \).

Substitution. Let \( t = \log_2 x \): \[ t^2 - 5t + 6 < 0 \quad \Longrightarrow \quad (t-2)(t-3) < 0 \quad \Longrightarrow \quad 2 < t < 3. \]

Returning to the original variable. \[ 2 < \log_2 x < 3 \quad \Longrightarrow \quad 4 < x < 8. \] Already in \( \mathcal{D} \).

Solution set: \( (4, 8) \).

Exercise 7. Solve \( \log_2 x + \log_4 x > 3 \).

Domain. \( \mathcal{D} = (0, +\infty) \).

Solution. \( \log_4 x = \dfrac{\log_2 x}{2} \). Let \( t = \log_2 x \): \[ t + \frac{t}{2} > 3 \quad \Longrightarrow \quad \frac{3t}{2} > 3 \quad \Longrightarrow \quad t > 2 \quad \Longrightarrow \quad \log_2 x > 2 \quad \Longrightarrow \quad x > 4. \] Already in \( \mathcal{D} \).

Solution set: \( (4, +\infty) \).

The graphical interpretation of logarithmic inequalities provides a qualitative picture of the solution set, complementing the analytic treatment.

Solving the inequality \( \log_a f(x) > k \) is geometrically equivalent to finding the values of the unknown for which the graph of \( y = \log_a f(x) \) lies above the horizontal line \( y = k \). If \( a > 1 \), the function is increasing and the graph rises above the threshold \( y = k \) for sufficiently large arguments; if \( 0 < a < 1 \), the function is decreasing and the graph rises above the threshold for sufficiently small (and positive) arguments. The graphical distinction between the two cases makes the mechanism of direction reversal visually clear.

Solving \( \log_a f(x) > \log_a g(x) \) is equivalent to finding the values of the unknown for which the graph of \( y = \log_a f(x) \) lies above the graph of \( y = \log_a g(x) \). The direction of the comparison depends on monotonicity: if \( a > 1 \), the graph of \( \log_a f(x) \) lies above that of \( \log_a g(x) \) exactly when \( f(x) > g(x) \); if \( 0 < a < 1 \), the same graph lies above exactly when \( f(x) < g(x) \), since the decreasing logarithm reverses the ordering of its arguments.

For inequalities solved by substitution, the solution set in \( t \) corresponds to a horizontal strip in the \( (x, t) \)-plane along the curve \( t = \log_a f(x) \); returning to the variable \( x \) amounts to determining the preimage of that strip under the function \( x \mapsto \log_a f(x) \), again taking into account the monotonicity with respect to the base.

The graphical interpretation also makes the role of the domain transparent: the branches of the graph exist only for \( x \in \mathcal{D} \), and any portion of a line or curve outside \( \mathcal{D} \) is simply absent from the Cartesian plane. Values excluded from the domain correspond to no point on the graph and therefore cannot constitute solutions of the original logarithmic inequality.