# Logarithmic Inequalities
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/logarithmic-inequalities/
Fetched from algebrica.org post 22097; source modified 2026-03-13T22:05:00.
## Introduction
Logarithmic inequalities are inequalities that involve one or more [logarithmic](../logarithms.md) expressions, in which the unknown \\(x\\) appears either in the argument of the logarithm or, in some cases, in the base itself. Before tackling logarithmic inequalities, it is essential to have a solid understanding of logarithms, their fundamental properties, and the standard methods used to solve [logarithmic equations](../logarithmic-inequalities.md), as these tools are crucial for approaching and resolving such inequalities correctly.
* * *
Logarithmic inequalities may have the general form:
\\[
\\log\_af(x) \\lesseqgtr \\log\_ag(x)
\\]
- \\(a\\) is the base of the logarithm, with (a > 0) and \\(a \\neq 1.\\)
- \\(f(x)\\) and \\(g(x)\\) are algebraic expressions depending on the variable \\(x.\\)
- The symbol \\(\\lesseqgtr\\) denotes one of the relations \\(\\le\\), \\(=\\), or \\(\\ge.\\)
Moreover, for the inequality to be well defined, the following conditions must be satisfied:
\\[
\\begin{cases} f(x) > 0\\[0.6em] g(x) > 0\\[0.6em] \\end{cases}
\\]
* * *
An inequality that contains a logarithmic expression in which the unknown variable does not appear in either the argument or the base of the logarithm is not a logarithmic inequality. In other words, an inequality such as:
\\[
\\log\_3(9) - 3x > 0
\\]
is not a logarithmic inequality, since the logarithmic term is a constant. By contrast,
\\[
\\log\_3(9x) - 3x > 0
\\]
is a logarithmic inequality, because the variable \\(x\\) appears inside the argument of the logarithm and directly affects its domain and behavior.
## How to solve logarithmic inequalities
The solution process for logarithmic inequalities is general, however, for explanatory convenience, let us consider the following inequality:
\\[
\\log\_a f(x) \\ge \\log\_a g(x)
\\]
The procedure can be structured into four fundamental steps:
- Determine the [domain](../determining-the-domain-of-a-function.md) of the inequality by imposing the admissibility conditions. The arguments of all logarithmic expressions must be strictly positive, and the base must satisfy:
\\[
\\begin{cases} f(x) > 0 \\[0.6em] g(x) > 0 \\[0.6em] \\end{cases}
\\]
- The base must satisfy:
\\[
\\begin{cases} a > 0 \\[0.6em] a \\neq 1 \\[0.6em] \\end{cases}
\\]
- If \\(a > 1\\), the logarithmic function is [increasing](../increasing-and-decreasing-functions.md) and the inequality preserves its direction when the logarithms are removed.
- If \\(0 < a < 1\\), the logarithmic function is decreasing and the direction of the inequality must be reversed when eliminating the logarithms.
- Use the monotonicity of the logarithmic function to remove the logarithms and reduce the problem to an equivalent algebraic inequality involving \\(f(x)\\) and \\(g(x)\\).
- Solve the resulting algebraic inequality and intersect the solution set with the domain previously determined, discarding any values that do not satisfy the original logarithmic conditions.
* * *
To explain the role played by the base of the logarithm, let us recall the behavior of the [logarithmic function](../logarithmic-function.md) when \\(0 < a < 1\\). From its graph, we observe that the function is strictly decreasing over its entire domain, with a vertical asymptote along the \\(y\\)-axis:
##### The dashed curve represents the logarithmic function with base \\(a > 1\\). In this case, the function is strictly increasing. In both cases, when \\(x = 1\\), the value of the logarithmic function is \\(0\\), and the graphs intersect at the point \\((1,0)\\).
## Example 1
Consider the logarithmic inequality:
\\[
\\log\_{\\frac{1}{2}}(x + 3) \\ge \\log\_{\\frac{1}{2}}(2x - 1)
\\]
We begin by determining the domain of the inequality. The arguments of the logarithms must be strictly positive:
\\[
\\begin{cases} x + 3 > 0 \\[0.6em] 2x - 1 > 0 \\end{cases}
\\]
Using a graphical representation and considering the solution intervals of the [linear inequalities](../linear-inequalities.md) in the previous system, we find that their intersection, which determines the domain of the logarithmic inequality, is precisely given by \\(x > 1/2\\).
\\[
-3
\\]
\\[
\\frac{1}{2}
\\]
Therefore, the domain \\(D\\) of the original inequality is given by the following interval:
\\[
\\left(-\\frac{1}{2}, +\\infty\\right)
\\]
* * *
Next, we analyze the base of the logarithm. Since the base satisfies
\\[
0 < \\frac{1}{2} < 1
\\]
the logarithmic function is strictly decreasing. As a consequence, when the logarithms are removed, the direction of the inequality must be reversed. Therefore, the given inequality:
\\[
\\log\_{\\frac{1}{2}}(x + 3) \\ge \\log\_{\\frac{1}{2}}(2x - 1)
\\]
is equivalent to:
\\[
x + 3 \\le 2x - 1
\\]
* * *
We now solve the resulting algebraic inequality:
\\[
x + 3 \\le 2x - 1 \\quad \\rightarrow \\quad x \\ge 4
\\]
Finally, we intersect this result with the domain previously determined. Since the domain requires \\(x > \\frac{1}{2}\\), the condition \\(x \\ge 4\\) is admissible.
Hence, the solution set of the logarithmic inequality is:
\\[
x \\ge 4
\\]
## Example 2
Consider the logarithmic inequality:
\\[
\\log\_{\\frac{1}{2}}(x+1) > \\log\_2(2-x)
\\]
We begin by determining the [domain](../determining-the-domain-of-a-function.md). The arguments of the logarithms must be strictly positive, hence:
\\[
\\begin{cases} x+1 > 0 \\[0.6em] 2-x > 0 \\[0.6em] \\end{cases}
\\]
Using a graphical representation and considering the solution intervals of the linear inequalities in the previous system, we find that their intersection is given by \\( -1 < x < 2\\).
\\[
-1
\\]
\\[
2
\\]
Therefore, the domain \\(D\\) of the original inequality is given by the following interval:
\\[
(-1, 2)
\\]
* * *
Next, we rewrite the logarithm with base \\(\\tfrac{1}{2}\\) in terms of base \\(2\\). Since \\(\\tfrac{1}{2} = 2^{-1}\\), we have
\\[
\\log\_{\\frac{1}{2}}(x+1) = \\frac{\\log\_2(x+1)}{\\log\_2(\\frac{1}{2})} = -\\log\_2(x+1)
\\]
Substituting into the original inequality, we obtain:
\\[
\\log\_2(x+1) < -\\log\_2(2-x)
\\]
Bringing all logarithmic terms to the same side and applying the properties of logarithms, we get:
\\[
\\log\_2(x+1) + \\log\_2(2-x) < 0
\\]
which, by the sum property of logarithms, stating that the [sum of two logarithms](../logarithms.md) equals the logarithm of their product, allows us to rewrite the expression as follows:
\\[
\\log\_2!\\bigl((x+1)(2-x)\\bigr) < 0
\\]
Since the logarithmic function with base \\(2>1\\) is strictly increasing, this inequality is equivalent to
\\[
0 < (x+1)(2-x) < 1
\\]
Within the domain \\((-1,2)\\), the product \\((x+1)(2-x)\\) is always positive, so it suffices to solve:
\\[
(x+1)(2-x) < 1
\\]
Expanding and simplifying, we obtain
\\[
x^2 - x - 1 > 0
\\]
The associated [quadratic equation](../quadratic-equations.md) has roots:
\\[
x = \\frac{1 \\pm \\sqrt{5}}{2}
\\]
The inequality is satisfied outside the interval determined by these roots. Intersecting this result with the domain \\((-1,2)\\), we finally obtain the solution set:
\\[
-1 < x < \\frac{1-\\sqrt{5}}{2}
\\]
