How do you solve log inequalities when base > 1?

Direction is preserved. For log_a(x) > k with a > 1, convert to x > a^k. For log_a(x) 0.

How do you solve log inequalities when base is between 0 and 1?

Direction reverses. For log_a(x) > k with 0 a^k. The decreasing function flips the inequality.

When do you flip the inequality sign in logarithmic inequalities?

Flip when the base is between 0 and 1 (like log_{1/2} or log_{0.1}). These logarithms are decreasing functions, so larger inputs give smaller outputs, reversing the inequality direction.

How do you compare two logarithms in an inequality?

For same base > 1: log_a(M) > log_a(N) means M > N. For same base between 0 and 1: log_a(M) > log_a(N) means M < N. Always check domain restrictions.

Why is domain important in log inequalities?

Every log argument must be positive. The algebraic solution must be intersected with domain restrictions. For log(x-3) > 1, you need x > 5 AND x > 3, giving x > 5.

How do you solve compound log inequalities?

Split into two inequalities, solve each, then find intersection. For 1 1 giving x > 2, and log₂(x) < 4 giving x < 16. Combined: 2 < x < 16.

How do you solve log inequalities with multiple logs?

Use log rules to combine into single logarithm first. For log(x) + log(x-2) > 3, combine to log(x(x-2)) > 3, then solve the resulting inequality and check domain.

Why does the base determine inequality direction?

Logarithms with base > 1 are increasing (larger x gives larger output), preserving direction. Base between 0 and 1 creates decreasing functions, reversing direction.

What is the solution to log₂(x) > 3?

Since base 2 > 1, direction is preserved: x > 2³ = 8. With domain x > 0, the solution is x > 8.

What is the solution to log_{1/2}(x) > 3?

Since base 1/2 0, the solution is 0 < x < 1/8.

Logarithmic Inequalities

Q: Why is domain important in log inequalities?

Every log argument must be positive. The algebraic solution must be intersected with domain restrictions. For log(x-3) > 1, you need x > 5 AND x > 3, giving x > 5.

Q: How do you solve log inequalities with multiple logs?

Use log rules to combine into single logarithm first. For log(x) + log(x-2) > 3, combine to log(x(x-2)) > 3, then solve the resulting inequality and check domain.

Key Terms

Base Greater Than One: Direction Preserved

Base Between Zero and One: Direction Reversed

Logarithms on Both Sides

Domain Considerations

Compound Inequalities

Inequalities with Combined Logarithms

Graphical Interpretation

Summary of Techniques

When Direction Depends on Base

Solving logarithmic inequalities follows similar algebraic steps as equations, with one critical addition: the base determines whether inequality direction is preserved or reversed. This behavior stems from the monotonicity property — logarithms with base greater than one are increasing functions, while those with base between zero and one are decreasing.

Key Terms

Inequality Concepts

Logarithmic Inequality— inequality direction depends on whether the base is greater than or less than

1

Monotonicity— the property that controls whether direction is preserved or reversed

Base (of a Logarithm)—

a > 1

preserves direction;

0 < a < 1

reverses it

Argument (of a Logarithm)— domain restriction: all arguments must remain positive

See All Algebra Definitions →

Base Greater Than One: Direction Preserved

When

a > 1

, the function

f(x) = \log_a(x)

is strictly increasing. Larger inputs produce larger outputs, so inequality direction is preserved when converting between logarithmic and exponential forms.

For

\log_a(x) > k

with

a > 1

\log_a(x) > k \implies x > a^k

For

\log_a(x) < k

with

a > 1

\log_a(x) < k \implies x < a^k

Example: Solve

\log_2(x) > 3

x > 2^3

x > 8

Combined with the domain requirement

x > 0

, the solution is

x > 8

.

Example: Solve

\log_5(x) \leq 2

x \leq 5^2

x \leq 25

With domain:

0 < x \leq 25

Base Between Zero and One: Direction Reversed

When

0 < a < 1

, the function

f(x) = \log_a(x)

is strictly decreasing. Larger inputs produce smaller outputs, so inequality direction reverses.

For

\log_a(x) > k

with

0 < a < 1

\log_a(x) > k \implies x < a^k

For

\log_a(x) < k

with

0 < a < 1

\log_a(x) < k \implies x > a^k

Example: Solve

\log_{1/2}(x) > 3

.

Since the base

1/2 < 1

, reverse the inequality:

x < \left(\frac{1}{2}\right)^3 = \frac{1}{8}

With domain:

0 < x < \frac{1}{8}

.

Example: Solve

\log_{0.1}(x) \leq -2

.

Reverse direction:

x \geq (0.1)^{-2} = 100

Solution:

x \geq 100

Logarithms on Both Sides

When the same base appears on both sides, use monotonicity to compare arguments directly.

For

a > 1

\log_a(M) > \log_a(N) \implies M > N

For

0 < a < 1

\log_a(M) > \log_a(N) \implies M < N

Example: Solve

\log_3(2x + 1) > \log_3(x + 4)

where base

3 > 1

2x + 1 > x + 4

x > 3

Check domain:

2x + 1 > 0

requires

x > -1/2

;

x + 4 > 0

requires

x > -4

. The intersection with

x > 3

is simply

x > 3

.

Example: Solve

\log_{1/3}(x - 1) < \log_{1/3}(5)

where base

1/3 < 1

.

Reverse when comparing arguments:

x - 1 > 5

x > 6

Domain requires

x - 1 > 0

, so

x > 1

. Final solution:

x > 6

Inequality form	Base a > 1 (direction preserved)	Base 0 < a < 1 (direction reversed)
logₐ(x) > k	x > aᵏ	x < aᵏ
logₐ(x) < k	x < aᵏ	x > aᵏ
logₐ(M) > logₐ(N)	M > N	M < N
logₐ(M) < logₐ(N)	M < N	M > N

Domain Considerations

Every logarithmic argument must be positive. This constraint intersects with the algebraic solution, often restricting the answer.

Example: Solve

\log_2(x - 3) > 1

.

Algebraically:

x - 3 > 2^1 = 2

x > 5

Domain:

x - 3 > 0 \implies x > 3

.

Intersection:

x > 5

satisfies

x > 3

, so the solution is

x > 5

.

Example: Solve

\log_4(x + 2) < 0

.

Algebraically:

x + 2 < 4^0 = 1

x < -1

Domain:

x + 2 > 0 \implies x > -2

.

Intersection:

-2 < x < -1

.

Failing to intersect with domain constraints is a common error. Always state domain restrictions at the start and combine them with the algebraic result.

Compound Inequalities

Compound inequalities bound the logarithm between two values.

Example: Solve

1 < \log_2(x) < 4

.

Split into two inequalities with base

2 > 1

\log_2(x) > 1 \implies x > 2

\log_2(x) < 4 \implies x < 16

Combined:

2 < x < 16

.

Example: Solve

-1 \leq \log_3(x + 1) \leq 2

\log_3(x+1) \geq -1 \implies x + 1 \geq 3^{-1} = \frac{1}{3} \implies x \geq -\frac{2}{3}

\log_3(x+1) \leq 2 \implies x + 1 \leq 9 \implies x \leq 8

Domain:

x + 1 > 0 \implies x > -1

.

Intersection:

-\frac{2}{3} \leq x \leq 8

already satisfies

x > -1

.

Solution:

-\frac{2}{3} \leq x \leq 8

Inequalities with Combined Logarithms

When logarithm rules are needed, condense first, then solve.

Example: Solve

\log_2(x) + \log_2(x - 2) > 3

.

Condense:

\log_2(x(x - 2)) > 3

\log_2(x^2 - 2x) > 3

Since base

2 > 1

x^2 - 2x > 8

x^2 - 2x - 8 > 0

(x - 4)(x + 2) > 0

This holds when

x < -2

x > 4

.

Domain:

x > 0

and

x - 2 > 0

gives

x > 2

.

Intersection with

x > 4

x < -2

: only

x > 4

survives.

Solution:

x > 4

Graphical Interpretation

The graph of

y = \log_a(x)

provides visual understanding of why direction depends on base.

For

a > 1

, the curve rises from left to right. The inequality

\log_a(x) > k

asks: where is the curve above the horizontal line

y = k

? Since the curve is increasing, this occurs to the right of the intersection point — for

x > a^k

.

For

0 < a < 1

, the curve falls from left to right. The inequality

\log_a(x) > k

asks the same question, but now the curve is above the line to the left of the intersection — for

x < a^k

.

The vertical asymptote at

x = 0

visually enforces the domain restriction. No portion of the graph exists for

x \leq 0

, so no solutions can come from that region.

Summary of Techniques

Solving logarithmic inequalities reduces to a small set of techniques selected by the form of the inequality. The base sets the direction, the form sets the conversion, and the domain trims the result. The table below collects each situation handled on this page with the technique and a worked example.

Situation	Technique	Worked example
Single log vs constant, base > 1	convert with direction preserved	log₂(x) > 3 → x > 8
Single log vs constant, 0 < base < 1	convert with direction reversed	log₍₁⁄₂₎(x) > 3 → x < 1⁄8
Logs on both sides, same base	drop the logs; reverse the comparison if base < 1	log₃(2x+1) > log₃(x+4) → x > 3
Composite argument	solve algebraically, then intersect with the domain (every argument > 0)	log₂(x−3) > 1 → x > 5
Compound (bounded)	split into two inequalities, solve each, intersect	1 < log₂(x) < 4 → 2 < x < 16
Multiple logs, same base	condense with logarithm rules first, then convert	log₂(x) + log₂(x−2) > 3 → x > 4
Visual check	read "curve above the line y = k" on the graph — direction follows whether the curve rises or falls	a > 1: solution to the right; 0 < a < 1: solution to the left