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Logarithmic Inequalities



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



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 11
Monotonicity— the property that controls whether direction is preserved or reversed
Base (of a Logarithm)— a>1a > 1 preserves direction; 0<a<10 < 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>1a > 1, the function f(x)=log⁡a(x)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\log_a(x) > k with a>1a > 1:
log⁡a(x)>k  ⟹  x>ak\log_a(x) > k \implies x > a^k


For log⁡a(x)<k\log_a(x) < k with a>1a > 1:
log⁡a(x)<k  ⟹  x<ak\log_a(x) < k \implies x < a^k


Example: Solve log⁡2(x)>3\log_2(x) > 3.
x>23x > 2^3

x>8x > 8


Combined with the domain requirement x>0x > 0, the solution is x>8x > 8.

Example: Solve log⁡5(x)≤2\log_5(x) \leq 2.
x≤52x \leq 5^2

x≤25x \leq 25


With domain: 0<x≤250 < x \leq 25.

Base Between Zero and One: Direction Reversed

When 0<a<10 < a < 1, the function f(x)=log⁡a(x)f(x) = \log_a(x) is strictly decreasing. Larger inputs produce smaller outputs, so inequality direction reverses.

For log⁡a(x)>k\log_a(x) > k with 0<a<10 < a < 1:
log⁡a(x)>k  ⟹  x<ak\log_a(x) > k \implies x < a^k


For log⁡a(x)<k\log_a(x) < k with 0<a<10 < a < 1:
log⁡a(x)<k  ⟹  x>ak\log_a(x) < k \implies x > a^k


Example: Solve log⁡1/2(x)>3\log_{1/2}(x) > 3.

Since the base 1/2<11/2 < 1, reverse the inequality:
x<(12)3=18x < \left(\frac{1}{2}\right)^3 = \frac{1}{8}


With domain: 0<x<180 < x < \frac{1}{8}.

Example: Solve log⁡0.1(x)≤−2\log_{0.1}(x) \leq -2.

Reverse direction:
x≥(0.1)−2=100x \geq (0.1)^{-2} = 100


Solution: x≥100x \geq 100.

Logarithms on Both Sides

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

For a>1a > 1:
log⁡a(M)>log⁡a(N)  ⟹  M>N\log_a(M) > \log_a(N) \implies M > N


For 0<a<10 < a < 1:
log⁡a(M)>log⁡a(N)  ⟹  M<N\log_a(M) > \log_a(N) \implies M < N


Example: Solve log⁡3(2x+1)>log⁡3(x+4)\log_3(2x + 1) > \log_3(x + 4) where base 3>13 > 1.
2x+1>x+42x + 1 > x + 4

x>3x > 3


Check domain: 2x+1>02x + 1 > 0 requires x>−1/2x > -1/2; x+4>0x + 4 > 0 requires x>−4x > -4. The intersection with x>3x > 3 is simply x>3x > 3.

Example: Solve log⁡1/3(x−1)<log⁡1/3(5)\log_{1/3}(x - 1) < \log_{1/3}(5) where base 1/3<11/3 < 1.

Reverse when comparing arguments:
x−1>5x - 1 > 5

x>6x > 6


Domain requires x−1>0x - 1 > 0, so x>1x > 1. Final solution: x>6x > 6.

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\log_2(x - 3) > 1.

Algebraically:
x−3>21=2x - 3 > 2^1 = 2

x>5x > 5


Domain: x−3>0  ⟹  x>3x - 3 > 0 \implies x > 3.

Intersection: x>5x > 5 satisfies x>3x > 3, so the solution is x>5x > 5.

Example: Solve log⁡4(x+2)<0\log_4(x + 2) < 0.

Algebraically:
x+2<40=1x + 2 < 4^0 = 1

x<−1x < -1


Domain: x+2>0  ⟹  x>−2x + 2 > 0 \implies x > -2.

Intersection: −2<x<−1-2 < 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)<41 < \log_2(x) < 4.

Split into two inequalities with base 2>12 > 1:
log⁡2(x)>1  ⟹  x>2\log_2(x) > 1 \implies x > 2

log⁡2(x)<4  ⟹  x<16\log_2(x) < 4 \implies x < 16


Combined: 2<x<162 < x < 16.

Example: Solve −1≤log⁡3(x+1)≤2-1 \leq \log_3(x + 1) \leq 2.

log⁡3(x+1)≥−1  ⟹  x+1≥3−1=13  ⟹  x≥−23\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)≤2  ⟹  x+1≤9  ⟹  x≤8\log_3(x+1) \leq 2 \implies x + 1 \leq 9 \implies x \leq 8


Domain: x+1>0  ⟹  x>−1x + 1 > 0 \implies x > -1.

Intersection: −23≤x≤8-\frac{2}{3} \leq x \leq 8 already satisfies x>−1x > -1.

Solution: −23≤x≤8-\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\log_2(x) + \log_2(x - 2) > 3.

Condense:
log⁡2(x(x−2))>3\log_2(x(x - 2)) > 3

log⁡2(x2−2x)>3\log_2(x^2 - 2x) > 3


Since base 2>12 > 1:
x2−2x>8x^2 - 2x > 8

x2−2x−8>0x^2 - 2x - 8 > 0

(x−4)(x+2)>0(x - 4)(x + 2) > 0


This holds when x<−2x < -2 or x>4x > 4.

Domain: x>0x > 0 and x−2>0x - 2 > 0 gives x>2x > 2.

Intersection with x>4x > 4 or x<−2x < -2: only x>4x > 4 survives.

Solution: x>4x > 4.

Graphical Interpretation

The graph of y=log⁡a(x)y = \log_a(x) provides visual understanding of why direction depends on base.

For a>1a > 1, the curve rises from left to right. The inequality log⁡a(x)>k\log_a(x) > k asks: where is the curve above the horizontal line y=ky = k? Since the curve is increasing, this occurs to the right of the intersection point — for x>akx > a^k.

For 0<a<10 < a < 1, the curve falls from left to right. The inequality log⁡a(x)>k\log_a(x) > k asks the same question, but now the curve is above the line to the left of the intersection — for x<akx < a^k.

The vertical asymptote at x=0x = 0 visually enforces the domain restriction. No portion of the graph exists for x≤0x \leq 0, so no solutions can come from that region.