Logarithmic Equations: Theory, Domain and Practice Problems

A logarithmic equation is an equation in which the unknown appears as the argument of at least one logarithm. Solving such equations requires not only proficiency in algebraic manipulation, but above all rigorous control of the existence conditions: every formally derived solution must be verified against the domain of the equation, otherwise extraneous solutions may be introduced..

Contents

• Review of the Logarithmic Function
• Domain of a Logarithmic Equation
• Definition and the Exponential Method
• Operational Properties of Logarithms
• Elementary Logarithmic Equations
• Equations Involving Sums or Differences of Logarithms
• Equations with Equal Logarithms
• Equations Involving Logarithms with Different Bases
• Equations Solvable by Substitution
• Extraneous Solutions: Analysis and Prevention
• General Solution Strategy
• Worked Examples
• Graphical Interpretation

Before tackling logarithmic equations, it is essential to recall with precision the fundamental properties of the logarithmic function, since the entire theory of their solution depends directly on these properties.

Fix \( a \in \mathbb{R} \) with \( a > 0 \) and \( a \neq 1 \). The exponential function \( e_a \colon \mathbb{R} \to (0, +\infty) \), defined by \( e_a(x) = a^x \), is strictly monotone — increasing when \( a > 1 \), decreasing when \( 0 < a < 1 \) — and therefore bijective onto its codomain \( (0, +\infty) \). Its inverse is the logarithmic function with base \( a \):

\[ \log_a \colon (0, +\infty) \longrightarrow \mathbb{R}, \qquad x \longmapsto \log_a x. \]

The natural domain of the logarithm is \( (0, +\infty) \): the logarithm of a non-positive number is not defined in \( \mathbb{R} \). This restriction is the source of all existence conditions in logarithmic equations.

The function \( x \mapsto \log_a x \) inherits strict monotonicity from the exponential function:

• it is strictly increasing when \( a > 1 \);
• it is strictly decreasing when \( 0 < a < 1 \).

Strict monotonicity implies injectivity: for every \( u, v \in (0, +\infty) \),

\[ \log_a u = \log_a v \quad \Longleftrightarrow \quad u = v. \]

This equivalence is the logical foundation of the method for solving equations with equal logarithms. We also recall the following standard values:

\[ \log_a 1 = 0, \qquad \log_a a = 1, \qquad \log_a a^k = k \quad \forall\, k \in \mathbb{R}. \]

The domain of a logarithmic equation is the set of real values of the unknown for which every expression appearing in the equation is well defined. Since the real logarithm is defined only for strictly positive arguments, for each term \( \log_a f_i(x) \), as \( i \) ranges over the index set of all logarithms present in the equation, one must impose the condition:

\[ f_i(x) > 0. \]

The domain of the equation is the intersection of all these conditions:

\[ \mathcal{D} = \bigcap_{i \in I} \bigl\{ x \in \mathbb{R} \mid f_i(x) > 0 \bigr\}, \]

where \( I \) is the index set of the logarithms present in the equation.

Fundamental rule. Once the formal solutions have been determined by algebraic transformations, only those belonging to \( \mathcal{D} \) may be retained. Values excluded from \( \mathcal{D} \) render at least one logarithm undefined and do not constitute solutions of the original equation, regardless of whether they satisfy the intermediate algebraic equations.

It is methodologically indispensable to determine \( \mathcal{D} \) before any algebraic manipulation: in this way one always has in mind the set within which solutions must be sought, and one avoids lending validity to steps that would presuppose the positivity of arguments without having established it.

The very definition of the logarithm provides the most direct solution method. For \( a > 0 \), \( a \neq 1 \), and \( x > 0 \), one has by definition:

\[ \log_a x = b \quad \Longleftrightarrow \quad a^b = x. \]

This equivalence allows the logarithmic equation \( \log_a f(x) = k \), with \( k \in \mathbb{R} \), to be transformed into the exponential equation \( f(x) = a^k \), which contains no logarithms. Note that the condition \( f(x) > 0 \) is automatically satisfied by every solution of \( f(x) = a^k \), since \( a^k > 0 \) for all \( k \in \mathbb{R} \) and every admissible base \( a \). This observation does not, however, exempt one from imposing and verifying the domain conditions established in \( \mathcal{D} \): they may involve other logarithms present in the original equation.

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

Domain. The only existence condition is \( x - 1 > 0 \), hence \( \mathcal{D} = (1, +\infty) \).

Solution. Applying the definition of logarithm:

\[ x - 1 = 3^2 = 9 \quad \Longrightarrow \quad x = 10. \]

Verification. \( 10 \in (1, +\infty) \). The solution is accepted.

Solution set: \( \{10\} \).

The following properties, valid for \( a > 0 \), \( a \neq 1 \), for all \( x, y > 0 \), and for all \( n \in \mathbb{R} \), follow directly from the corresponding properties of exponents:

\[ \log_a(xy) = \log_a x + \log_a y, \]

\[ \log_a\!\left(\frac{x}{y}\right) = \log_a x - \log_a y, \]

\[ \log_a(x^n) = n\log_a x. \]

These identities are the primary tools for reducing equations involving multiple logarithms to solvable canonical forms. Their application requires, however, scrupulous respect for the hypotheses of validity.

Critical warning. Each property is valid exclusively for strictly positive arguments:

• The property \( \log_a(xy) = \log_a x + \log_a y \) requires \( x > 0 \) and \( y > 0 \) separately. The product \( xy \) can be positive even when both factors are negative; in that case \( \log_a(xy) \) would be defined, but \( \log_a x \) and \( \log_a y \) would not. The identity therefore cannot be applied in reverse — from a product to a sum — without first guaranteeing the positivity of each factor.
• The property \( \log_a(x^n) = n\log_a x \) requires \( x > 0 \). A paradigmatic case is \( \log_a(x^2) = 2\log_a x \): the identity holds only for \( x > 0 \), whereas for \( x < 0 \) the left-hand side is defined (since \( x^2 > 0 \)) while the right-hand side is not.

We also recall the change-of-base formula: for every \( b > 0 \), \( b \neq 1 \), and every \( x > 0 \),

\[ \log_a x = \frac{\log_b x}{\log_b a}. \]

This formula is indispensable when an equation contains logarithms with different bases and one wishes to reduce them to a common base in order to apply the operational properties or the injectivity principle. Its systematic use is illustrated in Section §8.

A logarithmic equation is called elementary if it can be reduced, possibly after simple algebraic manipulation, to the canonical form:

\[ \log_a f(x) = k, \qquad k \in \mathbb{R}. \]

The solution method proceeds through the following steps:

1. Determine the domain: \( \mathcal{D} = \{x \in \mathbb{R} \mid f(x) > 0\} \).
2. Apply the definition of logarithm to obtain \( f(x) = a^k \).
3. Solve the equation \( f(x) = a^k \).
4. Verify that the solutions belong to \( \mathcal{D} \) and discard any extraneous solutions.

Example. Solve \( \log_{1/2}(3x - 5) = -3 \).

Domain. \( 3x - 5 > 0 \Rightarrow x > \tfrac{5}{3} \), hence \( \mathcal{D} = \bigl(\tfrac{5}{3}, +\infty\bigr) \).

Solution. Applying the definition:

\[ 3x - 5 = \left(\frac{1}{2}\right)^{-3} = 2^3 = 8 \quad \Longrightarrow \quad 3x = 13 \quad \Longrightarrow \quad x = \frac{13}{3}. \]

Verification. \( \tfrac{13}{3} \approx 4.33 > \tfrac{5}{3} \). Solution accepted.

Solution set: \( \Bigl\{\dfrac{13}{3}\Bigr\} \).

When an equation contains a sum or difference of logarithms with the same base, the product and quotient rules are applied to reduce the equation to a single logarithm, thereby bringing it to elementary form. As already noted, the domain conditions must be imposed on the original arguments, not on the argument of the logarithm produced by combining them.

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

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

Solution. Applying the product rule:

\[ \log_2[x(x-2)] = 3 \quad \Longrightarrow \quad x(x-2) = 2^3 = 8. \]

\[ x^2 - 2x - 8 = 0 \quad \Longrightarrow \quad (x-4)(x+2) = 0. \]

The formal solutions are \( x = 4 \) and \( x = -2 \).

Verification. \( x = 4 \in (2, +\infty) \): accepted. \( x = -2 \notin (2, +\infty) \): extraneous solution, discarded.

Solution set: \( \{4\} \).

Example 2. Solve \( \log_3(x+7) - \log_3(x-1) = 1 \).

Domain. \( x + 7 > 0 \) and \( x - 1 > 0 \), hence \( \mathcal{D} = (1, +\infty) \).

Solution. Applying the quotient rule:

\[ \log_3\!\left(\frac{x+7}{x-1}\right) = 1 \quad \Longrightarrow \quad \frac{x+7}{x-1} = 3. \]

\[ x + 7 = 3(x-1) \quad \Longrightarrow \quad x + 7 = 3x - 3 \quad \Longrightarrow \quad x = 5. \]

Verification. \( 5 \in (1, +\infty) \). Solution accepted.

Solution set: \( \{5\} \).

If the equation takes the form:

\[ \log_a f(x) = \log_a g(x), \]

the injectivity of the function \( t \mapsto \log_a t \) on \( (0, +\infty) \) allows one to assert that, for every \( u, v \in (0, +\infty) \):

\[ \log_a u = \log_a v \quad \Longleftrightarrow \quad u = v. \]

Provided \( f(x) > 0 \) and \( g(x) > 0 \), the logarithmic equation is therefore equivalent to the algebraic equation \( f(x) = g(x) \). A logical clarification is needed here, however: the conditions \( f(x) > 0 \) and \( g(x) > 0 \) are not equivalent in general, but they become so subject to the equality \( f(x) = g(x) \). More precisely: if \( x_0 \) satisfies \( f(x_0) = g(x_0) \), then \( f(x_0) > 0 \Longleftrightarrow g(x_0) > 0 \), and therefore verifying one of the two conditions suffices to guarantee both. From a methodological standpoint it is nevertheless more rigorous to impose both conditions a priori when determining \( \mathcal{D} \), without relying on this implication.

Example. Solve \( \log_5(x+1) = \log_5(2x-3) \).

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

Solution. By injectivity:

\[ x + 1 = 2x - 3 \quad \Longrightarrow \quad x = 4. \]

Verification. \( 4 > \tfrac{3}{2} \). Solution accepted.

Solution set: \( \{4\} \).

When an equation contains logarithms with different bases, neither the injectivity principle nor the operational properties can be applied directly. The standard method consists of reducing all logarithms to a common base by means of the change-of-base formula:

\[ \log_a x = \frac{\log_b x}{\log_b a}, \]

where the auxiliary base \( b \) is chosen so as to simplify the computations. The most common choices are \( b = 10 \) (common logarithm, denoted \( \lg \)) or \( b = e \) (natural logarithm, denoted \( \ln \)). In many cases, however, it is more convenient to take as the common base one of the bases already present in the equation.

Example 1. Solve \( \log_2 x + \log_4 x = 3 \).

Domain. Both logarithms require \( x > 0 \), hence \( \mathcal{D} = (0, +\infty) \).

Solution. We reduce \( \log_4 x \) to base \( 2 \) via the change-of-base formula:

\[ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}. \]

Substituting into the equation and setting \( t = \log_2 x \) for brevity:

\[ t + \frac{t}{2} = 3 \quad \Longrightarrow \quad \frac{3t}{2} = 3 \quad \Longrightarrow \quad t = 2. \]

Returning to the original variable: \( \log_2 x = 2 \Rightarrow x = 4 \).

Verification. \( 4 \in (0, +\infty) \). Solution accepted.

Solution set: \( \{4\} \).

Example 2. Solve \( \log_3 x \cdot \log_9 x = 4 \).

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

Solution. We reduce \( \log_9 x \) to base \( 3 \):

\[ \log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2}. \]

Setting \( t = \log_3 x \):

\[ t \cdot \frac{t}{2} = 4 \quad \Longrightarrow \quad \frac{t^2}{2} = 4 \quad \Longrightarrow \quad t^2 = 8 \quad \Longrightarrow \quad t = \pm 2\sqrt{2}. \]

Returning to the original variable:

\[ \log_3 x = 2\sqrt{2} \;\Rightarrow\; x = 3^{2\sqrt{2}}, \qquad \log_3 x = -2\sqrt{2} \;\Rightarrow\; x = 3^{-2\sqrt{2}}. \]

Verification. Both values are positive and belong to \( \mathcal{D} \). Both solutions are accepted.

Solution set: \( \bigl\{3^{-2\sqrt{2}},\; 3^{2\sqrt{2}}\bigr\} \).

An important class of logarithmic equations is that in which the logarithm appears as the argument of a polynomial expression. The typical form is:

\[ P\!\bigl(\log_a f(x)\bigr) = 0, \]

where \( P \) is a polynomial. The method consists of setting \( t = \log_a f(x) \), solving the algebraic equation \( P(t) = 0 \) in \( t \), and for each root \( t_k \) solving the resulting elementary equation \( \log_a f(x) = t_k \).

Warning. Each of the equations \( \log_a f(x) = t_k \) must be solved separately, with its own domain check. The substitution \( t = \log_a f(x) \) does not in itself introduce extraneous solutions, but the final step of returning to the original variable may do so, if one fails to verify that the solutions found belong to \( \mathcal{D} \).

Example 1. Solve \( (\log_2 x)^2 - 5\log_2 x + 6 = 0 \).

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

Substitution. Let \( t = \log_2 x \). The equation becomes:

\[ t^2 - 5t + 6 = 0 \quad \Longrightarrow \quad (t-2)(t-3) = 0 \quad \Longrightarrow \quad t = 2 \;\text{ or }\; t = 3. \]

Returning to the original variable.

\[ \log_2 x = 2 \;\Rightarrow\; x = 4, \qquad \log_2 x = 3 \;\Rightarrow\; x = 8. \]

Verification. \( 4, 8 \in (0, +\infty) \). Both solutions are accepted.

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

Example 2. Solve \( (\log_3 x)^2 - \log_3(x^4) + 3 = 0 \).

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

Simplification. For \( x > 0 \) the power rule applies: \( \log_3(x^4) = 4\log_3 x \). The equation becomes:

\[ (\log_3 x)^2 - 4\log_3 x + 3 = 0. \]

Substitution. Let \( t = \log_3 x \):

\[ t^2 - 4t + 3 = 0 \quad \Longrightarrow \quad (t-1)(t-3) = 0 \quad \Longrightarrow \quad t = 1 \;\text{ or }\; t = 3. \]

Returning to the original variable.

\[ \log_3 x = 1 \;\Rightarrow\; x = 3, \qquad \log_3 x = 3 \;\Rightarrow\; x = 27. \]

Verification. \( 3, 27 \in (0, +\infty) \). Both solutions are accepted.

Solution set: \( \{3, 27\} \).

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

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

Substitution. Let \( t = \log_5 x \):

\[ 2t^2 + 3t - 2 = 0 \quad \Longrightarrow \quad t = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}. \]

Hence \( t_1 = \tfrac{1}{2} \) and \( t_2 = -2 \).

Returning to the original variable.

\[ \log_5 x = \frac{1}{2} \;\Rightarrow\; x = 5^{1/2} = \sqrt{5}, \qquad \log_5 x = -2 \;\Rightarrow\; x = 5^{-2} = \frac{1}{25}. \]

Verification. \( \sqrt{5}, \tfrac{1}{25} \in (0, +\infty) \). Both solutions are accepted.

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

An extraneous solution is a value \( x_0 \) that satisfies an algebraic equation obtained in the course of the transformations, but does not belong to the domain \( \mathcal{D} \) of the original logarithmic equation. Such a value renders at least one logarithm undefined and is therefore not a valid solution.

Extraneous solutions typically arise in the following three circumstances.

1. Application of the product rule. Writing \( \log_a f(x) + \log_a g(x) = \log_a[f(x)g(x)] \) requires \( f(x) > 0 \) 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 equation, but the logarithms of the individual factors would not be defined in the original.
2. Application of the power rule. The identity \( \log_a[f(x)^n] = n\log_a f(x) \) is valid only when \( f(x) > 0 \). For \( f(x) < 0 \) and \( n \) even, the left-hand side is defined while the right-hand side is not: applying the identity therefore introduces inadmissible arguments.
3. Reduction to a polynomial equation. Solving a quadratic or higher-degree equation generally yields multiple roots; some of these may lie outside the domain \( \mathcal{D} \).

Illustrative example. Consider the equation:

\[ \log_2(x^2 - 5x + 6) = \log_2(x - 2) + 1. \]

Domain. The existence conditions are \( x^2 - 5x + 6 > 0 \) and \( x - 2 > 0 \). Factoring: \( (x-2)(x-3) > 0 \) when \( x < 2 \) or \( x > 3 \). Intersecting with \( x > 2 \) gives \( \mathcal{D} = (3, +\infty) \).

Solution. The right-hand side is rewritten using \( 1 = \log_2 2 \):

\[ \log_2(x^2 - 5x + 6) = \log_2[2(x-2)]. \]

By injectivity:

\[ x^2 - 5x + 6 = 2x - 4 \quad \Longrightarrow \quad x^2 - 7x + 10 = 0 \quad \Longrightarrow \quad (x-2)(x-5) = 0. \]

Formal solutions: \( x = 2 \) and \( x = 5 \).

Verification. \( x = 2 \notin (3, +\infty) \): extraneous solution, discarded. \( x = 5 \in (3, +\infty) \): accepted.

Solution set: \( \{5\} \).

The following scheme constitutes a complete and rigorous protocol applicable to every type of logarithmic equation treated in this article.

1. Determine the domain. For every logarithm \( \log_a f_i(x) \) in the equation, where \( i \) ranges over the index set \( I \) of all logarithms present, impose \( f_i(x) > 0 \). Compute \( \mathcal{D} = \bigcap_{i \in I} \{x \in \mathbb{R} \mid f_i(x) > 0\} \).
2. Reduce to a common base (if necessary). If the equation contains logarithms with different bases, apply the change-of-base formula to bring them to a uniform base.
3. Simplify using the properties of logarithms. Apply the product, quotient, and power rules — bearing in mind that they are valid only for positive arguments — to reduce the equation to one of the canonical forms: \( \log_a f(x) = k \), or \( \log_a f(x) = \log_a g(x) \), or \( P(\log_a f(x)) = 0 \).
4. Eliminate the logarithm. In the first form, pass to the exponential form \( f(x) = a^k \). In the second, exploit injectivity: \( f(x) = g(x) \). In the third, apply the substitution \( t = \log_a f(x) \) and solve the resulting algebraic equation in \( t \), then return to the original variable.
5. Solve the resulting algebraic equation.
6. Verify the domain conditions. Retain only the solutions belonging to \( \mathcal{D} \). Explicitly discard extraneous solutions, stating the reason for their exclusion.
7. State the solution set.

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

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

Solution. \( x + 3 = 2^4 = 16 \Rightarrow x = 13 \).

Verification. \( 13 \in (-3, +\infty) \). Solution accepted.

Solution set: \( \{13\} \).

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

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

Solution.

\[ \log_3[x(x-1)] = 1 \quad \Longrightarrow \quad x(x-1) = 3 \quad \Longrightarrow \quad x^2 - x - 3 = 0. \]

\[ x = \frac{1 \pm \sqrt{13}}{2}. \]

Verification. \( x_1 = \dfrac{1+\sqrt{13}}{2} \approx 2.30 > 1 \): accepted. \( x_2 = \dfrac{1-\sqrt{13}}{2} < 0 \): extraneous solution, discarded.

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

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

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

Solution. By injectivity: \( x + 1 = 2x - 3 \Rightarrow x = 4 \).

Verification. \( 4 > \tfrac{3}{2} \). Solution accepted.

Solution set: \( \{4\} \).

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.

\[ \log_2[(x-1)(x-5)] = 3 \quad \Longrightarrow \quad (x-1)(x-5) = 8. \]

\[ x^2 - 6x - 3 = 0 \quad \Longrightarrow \quad x = 3 \pm 2\sqrt{3}. \]

Verification. \( x_1 = 3 + 2\sqrt{3} \approx 6.46 > 5 \): accepted. \( x_2 = 3 - 2\sqrt{3} \approx -0.46 < 5 \): extraneous solution, discarded.

Solution set: \( \{3 + 2\sqrt{3}\} \).

Exercise 5. Solve \( \log_4(x^2 - 3x) = \log_4(x + 7) \).

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

Solution. By injectivity: \( x^2 - 3x = x + 7 \Rightarrow x^2 - 4x - 7 = 0 \Rightarrow x = 2 \pm \sqrt{11} \).

Verification. \( x_1 = 2+\sqrt{11} \approx 5.32 \in (3, +\infty) \): accepted. \( x_2 = 2-\sqrt{11} \approx -1.32 \in (-7, 0) \): accepted.

Solution set: \( \{2-\sqrt{11},\; 2+\sqrt{11}\} \).

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

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

Solution. Let \( t = \log_2 x \):

\[ t^2 - 5t + 6 = 0 \quad \Longrightarrow \quad t = 2 \;\text{ or }\; t = 3. \]

\[ \log_2 x = 2 \Rightarrow x = 4, \qquad \log_2 x = 3 \Rightarrow x = 8. \]

Verification. \( 4, 8 \in (0, +\infty) \). Both accepted.

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 \( t = \log_2 x \):

\[ t + \frac{t}{2} = 3 \quad \Longrightarrow \quad t = 2 \quad \Longrightarrow \quad x = 4. \]

Verification. \( 4 \in (0, +\infty) \). Solution accepted.

Solution set: \( \{4\} \).

The graphical interpretation of logarithmic equations provides a qualitative picture of the number and location of solutions, complementing the analytical treatment.

Solving the equation \( \log_a f(x) = k \) is geometrically equivalent to finding the \( x \)-coordinates of the intersection points between the graph of \( y = \log_a f(x) \) and the horizontal line \( y = k \). Since the logarithmic function is strictly monotone, on each connected subinterval of the domain over which \( f \) is itself strictly monotone, the composition \( \log_a \circ f \) is also strictly monotone, and therefore the equation has at most one solution on each such interval. This fact allows one to establish an upper bound on the number of solutions a priori, before carrying out any computation.

Solving \( \log_a f(x) = \log_a g(x) \) is equivalent to finding the points at which the graphs of \( y = \log_a f(x) \) and \( y = \log_a g(x) \) intersect. Such points must lie in the common domain \( \mathcal{D}_f \cap \mathcal{D}_g \) of the two functions; any apparent crossings outside this domain do not correspond to solutions of the original equation.

The graphical interpretation also makes clear why extraneous solutions are not solutions: they correspond to values of the unknown for which one or more branches of the graph simply do not exist, since the logarithmic function is not defined outside \( (0, +\infty) \). An extraneous solution is not a point on the graph — it is an algebraic artefact devoid of geometric content within the original equation.