# Logarithmic Inequalities

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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:

![Graph of the logarithmic function with base between zero and one.](https://algebrica.org/wp-content/uploads/resources/images/logharithm-5-1.png "Graph of the logarithmic function with base between zero and one.")

##### 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}
\\]