Domain & Range in Logarithmic Inequalities: A Common JEE Pitfall
Master complex inequalities like $\log_x(2x-1) > 0$ where the base itself is variable. Step-by-step case analysis approach.
Why Variable Base Logarithms Are Tricky
When the base of a logarithm is variable, you need to consider multiple cases because the behavior changes dramatically based on whether the base is between 0 and 1 or greater than 1.
⚠️ The Critical Insight Most Students Miss
For $\log_a b > 0$, the inequality direction depends on the base $a$:
- If $a > 1$: $\log_a b > 0 \Rightarrow b > 1$
- If $0 < a < 1$: $\log_a b > 0 \Rightarrow 0 < b < 1$ (inequality reverses!)
🎯 Problem Navigation
Problem 1: Basic Case Analysis
Solve for $x$: $\log_x(2x-1) > 0$
✅ Systematic Solution
📌 Step 1: Fundamental Conditions
1 Base condition: $x > 0$ and $x \neq 1$
2 Argument condition: $2x - 1 > 0 \Rightarrow x > \frac{1}{2}$
Initial domain: $x > \frac{1}{2}$ and $x \neq 1$
📌 Step 2: Case Analysis Based on Base
Case 1: $x > 1$
For $x > 1$, $\log_x(2x-1) > 0 \Rightarrow 2x - 1 > 1$
$2x > 2 \Rightarrow x > 1$
Solution for Case 1: $x > 1$
Case 2: $0 < x < 1$
For $0 < x < 1$, $\log_x(2x-1) > 0 \Rightarrow 0 < 2x - 1 < 1$
$1 < 2x < 2 \Rightarrow \frac{1}{2} < x < 1$
Solution for Case 2: $\frac{1}{2} < x < 1$
📌 Step 3: Combine Solutions
From Case 1: $x > 1$
From Case 2: $\frac{1}{2} < x < 1$
Final Solution: $\left(\frac{1}{2}, 1\right) \cup (1, \infty)$
💡 Key Learning
- Always split into two cases: base > 1 and 0 < base < 1
- Remember: Inequality reverses when base is between 0 and 1
- Don't forget the fundamental conditions: base > 0, base ≠ 1, argument > 0
Problem 2: Complex Argument
Solve for $x$: $\log_x(x^2 + x - 6) < 2$
✅ Systematic Solution
📌 Step 1: Fundamental Conditions
1 Base: $x > 0$, $x \neq 1$
2 Argument: $x^2 + x - 6 > 0$
Factor: $(x+3)(x-2) > 0 \Rightarrow x < -3$ or $x > 2$
Initial domain: $x > 2$ (combining with $x > 0$)
📌 Step 2: Rewrite Inequality
$\log_x(x^2 + x - 6) < 2$
Using logarithm property: $\log_x(x^2 + x - 6) < \log_x(x^2)$
Note: We use $x^2$ because $\log_x(x^2) = 2$
📌 Step 3: Case Analysis
Case 1: $x > 1$
Inequality direction preserved: $x^2 + x - 6 < x^2$
$x - 6 < 0 \Rightarrow x < 6$
Combine with $x > 2$ and $x > 1$: $2 < x < 6$
Case 2: $0 < x < 1$
Inequality reverses: $x^2 + x - 6 > x^2$
$x - 6 > 0 \Rightarrow x > 6$
But $x > 6$ contradicts $0 < x < 1$
No solution in this case
📌 Step 4: Final Solution
From Case 1: $2 < x < 6$
From Case 2: No solution
Final Solution: $(2, 6)$
📋 Quick Reference: Inequality Behavior
When Base $a > 1$:
- $\log_a f(x) > \log_a g(x) \Rightarrow f(x) > g(x)$
- $\log_a f(x) < \log_a g(x) \Rightarrow f(x) < g(x)$
- Function is increasing
- Inequality direction preserved
When $0 < a < 1$:
- $\log_a f(x) > \log_a g(x) \Rightarrow f(x) < g(x)$
- $\log_a f(x) < \log_a g(x) \Rightarrow f(x) > g(x)$
- Function is decreasing
- Inequality direction reversed
🎯 Memory Aid: "Small base (0
Problem 3: Multiple Logarithmic Conditions
Find domain of $f(x) = \sqrt{\log_x(x^2 - 3x + 2)}$
✅ Systematic Solution
📌 Step 1: Square Root Condition
For $\sqrt{\log_x(x^2 - 3x + 2)}$ to be defined:
$\log_x(x^2 - 3x + 2) \geq 0$
📌 Step 2: Fundamental Conditions
1 Base: $x > 0$, $x \neq 1$
2 Argument: $x^2 - 3x + 2 > 0$
Factor: $(x-1)(x-2) > 0 \Rightarrow x < 1$ or $x > 2$
📌 Step 3: Solve $\log_x(x^2 - 3x + 2) \geq 0$
Case 1: $x > 1$
$\log_x(x^2 - 3x + 2) \geq 0 \Rightarrow x^2 - 3x + 2 \geq 1$
$x^2 - 3x + 1 \geq 0$
Roots: $x = \frac{3 \pm \sqrt{5}}{2} \approx 0.38, 2.62$
Solution: $x \leq \frac{3 - \sqrt{5}}{2}$ or $x \geq \frac{3 + \sqrt{5}}{2}$
Combine with $x > 1$: $x \geq \frac{3 + \sqrt{5}}{2}$
Case 2: $0 < x < 1$
$\log_x(x^2 - 3x + 2) \geq 0 \Rightarrow 0 < x^2 - 3x + 2 \leq 1$
From argument condition: $x^2 - 3x + 2 > 0$ already gives $x < 1$
And $x^2 - 3x + 2 \leq 1 \Rightarrow x^2 - 3x + 1 \leq 0$
Solution: $\frac{3 - \sqrt{5}}{2} \leq x \leq \frac{3 + \sqrt{5}}{2}$
Combine with $0 < x < 1$: $\frac{3 - \sqrt{5}}{2} \leq x < 1$
📌 Step 4: Final Domain
From Case 1: $x \geq \frac{3 + \sqrt{5}}{2}$
From Case 2: $\frac{3 - \sqrt{5}}{2} \leq x < 1$
Final Domain: $\left[\frac{3 - \sqrt{5}}{2}, 1\right) \cup \left[\frac{3 + \sqrt{5}}{2}, \infty\right)$
🔄 Systematic Approach for Variable Base Logarithms
Identify Fundamental Conditions
Base > 0, Base ≠ 1, Argument > 0
Split into Two Cases
Case 1: Base > 1, Case 2: 0 < Base < 1
Apply Correct Inequality Rules
Preserve direction for base > 1, reverse for 0 < base < 1
Solve Each Case Separately
Combine with fundamental conditions
Take Union of Valid Solutions
Combine solutions from both cases
Problems 4-6 Available in Full Version
Includes quadratic bases, nested logarithms, and advanced composite functions
🎯 Test Your Understanding
Try these problems using the systematic approach:
1. Solve: $\log_x(3x-2) \leq 1$
2. Find domain: $f(x) = \sqrt{\log_x(4-x)}$
3. Solve: $\log_{x^2}(x+1) > \frac{1}{2}$
🚫 Common Pitfalls to Avoid
Forgetting Case Analysis
Solving only for base > 1 and missing the 0 < base < 1 case
Ignoring Fundamental Conditions
Not checking base > 0, base ≠ 1, argument > 0
Wrong Inequality Direction
Not reversing inequality when base is between 0 and 1
Incorrect Domain Combination
Taking intersection instead of union of cases
Master Variable Base Logarithms!
These problems seem tricky but become routine with the systematic case analysis approach