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Logarithmic Inequality Solver

How to use Logarithmic Inequality Solver
  1. Build the inequality in the display at the top: click a Log button for ln, log or a base, then the argument and a closing bracket, then one of the four Ineq symbols and the rest with the Var, Num and Op buttons, or type on the keyboard. Learn more about building inequalities
  2. Clear empties the display, the curved arrow undoes the last edit and the ⌫ button deletes the character before the caret. Learn more about editing
  3. Click Solve, or press Enter in the display, once the inequality is complete; an expression without an inequality symbol or without a logarithm of the variable is refused. Learn more about solving
  4. Open Try an Example and click a card to load a random inequality of that form. Learn more about the templates
  5. The right panel lists the steps from isolating the logarithm to the exponential form and the domain, and the answer card gives the inequality and the interval; the box under the solver explains the method. Learn more about the solution
  6. The argument of every logarithm must stay positive, so a less-than answer becomes a bounded interval and an impossible one becomes No solution. Learn more about the domain
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Solve inequalities with logarithms step by step

Enter inequality...
Log
Var
Num
Op
Ineq

Select an inequality type or enter your own, then click Solve to see the step-by-step solution.

How the solver works

The solver reads one inequality in xx, yy or nn with the variable inside a logarithm, in base ee, 1010 or any positive base other than 11. When both sides are logarithms with the same base it compares the arguments; otherwise it isolates the logarithm, rewrites logb(A)log_b(A) compared with cc as AA compared with bcb^c, flipping the sign when the base is below 11, solves the linear inequality that remains and intersects the result with the domain where the argument is positive. The answer is shown as an inequality and as an interval. Learn more about logarithmic inequalities





Getting Started

Demolog₂(x) > 3, Solve, Clear, the domain at work
Step 0 of 4
The solver shows an inequality display at the top with a blinking yellow caret marking the cursor position. Click anywhere in the display to place the cursor, or use arrow keys, Home, and End for keyboard navigation. Type directly or use the button panel below.

Three quick experiments:

• Type log⁡2(x)>3\log_2(x) > 3 and press Enter — the right panel converts to exponential form x>23=8x > 2^3 = 8, applies the domain x>0x > 0 (automatically satisfied since 8>08 > 0), and reports x>8x > 8.
• Click the "Linear Argument" example template — an inequality like log⁡2(2x+1)>3\log_2(2x + 1) > 3 loads. The solver converts to 2x+1>82x + 1 > 8, solves the linear inequality, then intersects with the domain 2x+1>02x + 1 > 0.
• Try the "Negative Right Side" template — an inequality such as log⁡2(x)>−3\log_2(x) > -3. Converting gives x>2−3=1/8x > 2^{-3} = 1/8, automatically positive so the domain is satisfied.

The Solve button is disabled until you enter something. The dedicated "Log" row inserts ln⁡\ln, log⁡\log (base 10), log⁡2\log_2, log⁡3\log_3, and log⁡5\log_5 as single button clicks. An expression without an inequality symbol is refused with "Input must be an inequality (use <, >, ≤, or ≥)", one without a logarithm of the variable with "No logarithm with variable found in inequality", and different bases on the two sides are not yet supported. log⁡2(x)<3\log_2(x) < 3 converts to x<8x < 8 and, with the domain x>0x > 0, gives (0,8)(0, 8). What makes an inequality logarithmic is explained in what is a logarithmic inequality.

Building Inequalities with Buttons and Keyboard

DemoA linear argument, undo, delete, Solve
Step 0 of 4
The button panel groups inputs by type. All inputs can be made either by clicking or typing on the keyboard.

• Log row — one click inserts the logarithm function with the appropriate base: ln⁡\ln (natural log), log⁡\log (base 10), log⁡2\log_2, log⁡3\log_3, log⁡5\log_5. After insertion the cursor sits inside the parentheses, ready for the argument.
• Variable row — xx, yy, nn.
• Number row — digits 0 through 9 and the decimal point.
• Operator row — plus, minus, multiplication, division, caret.
• Ineq row — the four inequality symbols.
• Special row — ee for the natural exponential base, and parentheses.

For non-standard bases, type log⁡\log then underscore then the base number then open parenthesis: log⁡_7(x)\log\_7(x) for log⁡7(x)\log_7(x). The parser recognizes both the underscore-base form and the bare log⁡\log (default base 10) and ln⁡\ln (base ee) forms.

Keyboard shortcuts: type letters and digits directly, use star or slash for multiplication and division, type << and >> directly, type <=<= or >=>= to insert ≤\leq and ≥\geq, and press Enter to solve. Ctrl+Z undoes up to fifty edits back. The Clear button empties the display entirely; the curved-arrow button steps through the undo stack one edit at a time; the ⌫ button deletes the character before the caret. log⁡2(2x+1)>3\log_2(2x + 1) > 3 is typed by clicking the base button, then 2x+12x + 1 and the closing bracket; it converts to 2x+1>82x + 1 > 8 and gives x>3.5x > 3.5. Why the domain must be checked is explained in domain and direction.

Try an Example — Eight Form Templates

DemoOpen the templates, load two forms
Step 0 of 4
Click the "Try an Example" header to expand the template panel. Each card generates a random inequality of that form. Clicking again produces a new random version.

• Simple — log⁡b(x)>c\log_b(x) > c with b∈{2,3,5,10}b \in \{2, 3, 5, 10\}. The base case with a single-variable argument.
• Natural Log — ln⁡(x)>c\ln(x) > c. Uses the natural base ee, which produces clean exponential-form right sides.
• With Coefficient — a⋅log⁡b(x)>ca \cdot \log_b(x) > c. Requires dividing by aa to isolate the logarithm before converting.
• Linear Argument — log⁡b(mx+n)>c\log_b(mx + n) > c. The logarithm argument is itself a linear expression; the post-conversion linear inequality is non-trivial.
• With Constant — log⁡b(x)+c>d\log_b(x) + c > d. Requires subtracting cc to isolate the logarithm.
• Natural Linear — ln⁡(mx+n)>c\ln(mx + n) > c. Combines the natural base with a linear argument.
• Same Base Both Sides — log⁡b(f(x))>log⁡b(c)\log_b(f(x)) > \log_b(c). Exercises the same-base comparison shortcut: the arguments are compared directly.
• Negative Right Side — log⁡b(x)>−c\log_b(x) > -c. Tests the case where the exponential value b−cb^{-c} is a small positive number rather than zero or negative.

Roughly 80 percent of generated inequalities use clean bases and right-hand sides that produce exact integer or rational boundary values. Loading a template clears the previous solution; click Solve to work the new inequality. The five stages applied to every template are listed in the solving process explained.

Reading the Step-by-Step Solution

DemoA coefficient to divide out, then the natural log
Step 0 of 3
The solution panel shows each algebraic move as a labeled step. The exact sequence depends on the structural case.

• Rearrange Inequality (when applicable) — if the logarithm is on the right, the solver swaps sides and flips the direction.
• Evaluate Constant — simplifies the non-logarithm side to a single number.
• Isolate Logarithm — subtracts additive constants or divides by a coefficient outside the logarithm to get the form log⁡b(expr)□c\log_b(\text{expr}) \square c.
• Compare Logarithms (for same-base case) — if both sides are logarithms with the same base, the solver removes the logarithm from both sides and continues with the arguments. Direction is preserved when the base is greater than 1 and flipped when less than 1.
• Convert to Exponential Form — restates log⁡b(A)□c\log_b(A) \square c as A□bcA \square b^c. Direction flips when the base is less than 1.
• Evaluate Exponential — computes bcb^c to a single number.
• Domain Constraint — reminds that the original argument must be positive for the logarithm to be defined.
• Solve Linear Inequality — finishes the post-conversion linear inequality in the variable.
• Apply Domain Constraint — intersects the algebraic solution with the requirement that the argument be positive. May produce a bounded interval rather than a single ray.

The final-answer card shows the result as an inequality form plus interval notation. 2⋅log⁡(x)≤42 \cdot \log(x) \leq 4 is divided by 22 first, converts to x≤100x \leq 100 and, with the domain, gives (0,100](0, 100]; ln⁡(x)≥0\ln(x) \geq 0 gives x≥1x \geq 1, the interval [1,∞)[1, \infty). The box under the solver, How the solver works, summarises the method in a few lines and links to the theory below. The two things that can go wrong are explained in domain and direction.

Domain and Direction — The Two Hazards

Logarithmic inequality solving has two distinct hazards. Tracking both is critical.

Hazard 1: Domain restriction. A logarithm is defined only when its argument is positive. So log⁡b(f(x))\log_b(f(x)) is defined when f(x)>0f(x) > 0. Any candidate value of xx that violates this is discarded, regardless of what the algebraic solution says.

• For an isolated log⁡b(x)□c\log_b(x) \square c, the post-conversion inequality x□bcx \square b^c may include values where x≤0x \leq 0 that must be excluded.
• For log⁡b(linear expression)\log_b(\text{linear expression}), the argument must be positive: mx+n>0mx + n > 0 adds its own boundary to combine with the algebraic answer.
• The final answer is always the intersection: the algebraic solution AND the domain.

Hazard 2: Direction flip with bases less than 1. The logarithm function log⁡b\log_b is increasing when b>1b > 1 and decreasing when 0<b<10 < b < 1. Converting between logarithmic and exponential forms with a base less than 1 reverses the inequality direction.

• log⁡1/2(x)>3\log_{1/2}(x) > 3 converts to x<(1/2)3=1/8x < (1/2)^3 = 1/8 (direction flipped).
• log⁡2(x)>3\log_2(x) > 3 converts to x>23=8x > 2^3 = 8 (direction preserved).

The combination. When both hazards apply, the solver handles them sequentially: first the algebraic conversion with the appropriate direction, then the domain intersection. The result is either a single ray, a bounded interval, or no solution.

Examples.

• log⁡2(x)<3\log_2(x) < 3 becomes x<8x < 8 with domain x>0x > 0. Intersection: 0<x<80 < x < 8, a bounded interval.
• log⁡2(x)>−3\log_2(x) > -3 becomes x>1/8x > 1/8 with domain x>0x > 0. Domain is automatic since 1/8>01/8 > 0. Final answer: x>1/8x > 1/8.
• log⁡2(x)<−3\log_2(x) < -3 becomes x<1/8x < 1/8 with domain x>0x > 0. Intersection: 0<x<1/80 < x < 1/8.

For deeper coverage see logarithm and domain of a function.

The domain hazard is treated in domain considerations; the direction rule in base greater than one: direction preserved and base between zero and one: direction reversed.

What is a Logarithmic Inequality?

A logarithmic inequality is an inequality where the variable appears inside a logarithm. The standard form is

log⁡b(f(x))  □  c\log_b(f(x)) \;\square\; c


where b>0b > 0 and b≠1b \neq 1 is the base, f(x)f(x) is a function of the variable (usually linear), cc is a real number, and □\square is one of <<, >>, ≤\leq, ≥\geq.

The two regimes by base.

• b>1b > 1: log⁡b\log_b is strictly increasing. Larger argument means larger logarithm. Converting between logarithmic and exponential form preserves direction.
• 0<b<10 < b < 1: log⁡b\log_b is strictly decreasing. Larger argument means smaller logarithm. Conversion reverses direction.

The domain restriction. For any base, log⁡b(f(x))\log_b(f(x)) requires f(x)>0f(x) > 0. This restriction must be enforced separately and intersected with the algebraic solution.

Solution shapes. Depending on the structural case, a logarithmic inequality can produce:

• A single ray (most common when the algebraic solution agrees with the domain).
• A bounded interval (when the algebraic solution caps from above and the domain caps from below).
• A half-line starting at the domain boundary.
• No solution (when the algebraic solution is disjoint from the domain).

The pivotal move. Converting log⁡b(f(x))□c\log_b(f(x)) \square c to f(x)□bcf(x) \square b^c extracts the variable from the logarithm. From there, standard linear (or polynomial) inequality techniques finish the problem.

Same-base shortcut. When both sides are logarithms with the same base, the conversion can be skipped: log⁡b(A)□log⁡b(B)\log_b(A) \square \log_b(B) becomes A□BA \square B directly, with direction reversed for bases less than 1. Both arguments must still be positive.

For deeper coverage see logarithmic inequality, logarithm, and exponential function.

The place of these inequalities among the logarithm topics is logarithmic inequalities, the picture behind the solution shapes is graphical interpretation, and the allowed inputs are restrictions on base and argument.

The Solving Process Explained

Five stages handle any standard logarithmic inequality.

• Stage 1: Isolate the logarithm. Move the logarithm to one side. Subtract any constants added to it. Divide by any coefficient multiplying it (flipping direction if the coefficient is negative). The goal is to reach log⁡b(f(x))□c\log_b(f(x)) \square c where the right side is a single number.

• Stage 2: Recognize same-base shortcut. If both sides are logarithms with the same base after isolation, skip the exponential conversion and compare arguments directly: log⁡b(A)□log⁡b(B)\log_b(A) \square \log_b(B) becomes A□BA \square B (or flipped if b<1b < 1). Both A>0A > 0 and B>0B > 0 must hold.

• Stage 3: Convert to exponential form. Replace log⁡b(f(x))□c\log_b(f(x)) \square c with f(x)□bcf(x) \square b^c. Direction flips if and only if b<1b < 1.

• Stage 4: Solve the resulting inequality. Compute bcb^c as a number. The inequality is now f(x)□bcf(x) \square b^c. If f(x)f(x) is linear, the standard linear-inequality moves finish it.

• Stage 5: Intersect with the domain. The original argument f(x)f(x) must be positive. Solve f(x)>0f(x) > 0 and intersect with the result of stage 4. The intersection is the final answer.

The implementation. Internally, the solver navigates this decision tree based on the structure of the input AST. It detects same-base shortcuts, applies appropriate direction-flip logic per base, and assembles the final answer by intersecting the algebraic solution with the domain. The result is rendered as an inequality plus the corresponding interval notation.

Edge cases the solver handles automatically.

• No-solution after intersection — when the algebraic answer and the domain do not overlap.
• Domain automatic — when bc>0b^c > 0 trivially satisfies the domain (which is always true since b>0b > 0).
• Bounded interval — when the algebraic answer caps from above and the domain caps from below (or vice versa).

For comprehensive treatment see solving logarithmic inequalities and logarithm rules.

The same-base shortcut is logarithms on both sides, merging several logarithms first is inequalities with combined logarithms, and the whole procedure is summarised in summary of techniques.