Logarithmic Inequalities: JEE Main Math Guide

Logarithmic inequalities combine two challenge points: the base-dependent reversal of inequality direction, and the domain restrictions that must always be applied first. JEE Main tests them in one to two questions per session, often in the context of finding the set of values of x satisfying a log inequality. Mistakes are almost always one of two types — forgetting to flip the inequality when the base < 1, or forgetting the domain condition. This guide eliminates both.

The Core Rule: Base-Dependent Direction

If b > 1 (base greater than 1): log_b(f(x)) > log_b(g(x)) ⟺ f(x) > g(x) (inequality direction preserved). If 0 < b < 1 (base between 0 and 1): log_b(f(x)) > log_b(g(x)) ⟺ f(x) < g(x) (inequality direction reverses). This is because log is an increasing function when b > 1 and a decreasing function when 0 < b < 1. Always check the base first before writing any inequality in terms of the argument.

Domain condition: for any logarithm to be defined, the argument must be strictly positive (> 0). When solving log inequalities, first write down the domain (argument > 0), then solve the transformed inequality, then take the intersection. Forgetting the domain is the most common error — JEE often designs the correct answer to require intersecting with the domain. Take a free algebra mock. See our logarithms guide and our inequalities guide.

Worked Example: Variable Base

Solve: log_{x}(x² − 3x + 2) > 1 (where x is the base). Two cases based on base: Case 1 (x > 1): inequality direction preserved: x² − 3x + 2 > x¹ = x → x² − 4x + 2 > 0 → x < 2 − √2 or x > 2 + √2. Combined with x > 1: x > 2 + √2. Also need domain: x² − 3x + 2 > 0 → (x−1)(x−2) > 0 → x < 1 or x > 2. Combined with x > 1: x > 2. And x > 2 + √2 AND x > 2 gives x > 2 + √2. Case 2 (0 < x < 1, x ≠ 1): inequality direction reverses: x² − 3x + 2 < x → x² − 4x + 2 < 0 → 2 − √2 < x < 2 + √2. Combined with 0 < x < 1: 0 < x < 1 (since 2 − √2 ≈ 0.586 < 1). Also domain x < 1 or x > 2. Combined: 0 < x < 1. Intersection: 0.586 < x < 1. Full solution: (2−√2, 1) ∪ (2+√2, ∞) — but also x ≠ 1 and x ≠ base constraints (x > 0, x ≠ 1).

Chain Log Inequalities and Composition

For log(log(x)) > 0: domain first: log(x) must be defined → x > 0, and log(x) must be positive (since log of argument must be > 0 for the outer log to be defined) → x > 1 (assuming log base 10 or base e with the right sense). Then: log(log(x)) > 0 → log(x) > 1 (base 10) → x > 10. Final answer: x > 10. For log(log(x)) to be defined: x > 1 (from inner log > 0 requirement). So domain: x > 1, and solution: x > 10. Combining: x > 10.

Logarithmic Inequality with Products and Quotients

For log(f(x) × g(x)) > 0 where f and g might individually be negative: the log is defined only when f×g > 0. Then log(f×g) > 0 → f×g > 1 (base e) or f×g > 10 (base 10). The challenge: f×g > 0 means either both positive (f > 0, g > 0) or both negative (f < 0, g < 0). And f×g > 1 requires the product to exceed 1. Work through each sub-case systematically. JEE question: find x such that log((x+2)(x−5)) > log(2). Base: the log base must be specified (assume base 10 by default). Direction preserved (base > 1): (x+2)(x−5) > 2 AND (x+2)(x−5) > 0 (domain). First: domain → x < −2 or x > 5. Inequality: x² − 3x − 10 > 2 → x² − 3x − 12 > 0 → x < (3−√57)/2 or x > (3+√57)/2. Intersect with domain: x > (3+√57)/2 ≈ 5.27, i.e., x > (3+√57)/2. For domain of logarithms and related function types see our function types and properties guide.