# EXAMPLE OF SOLVING LOGARITHMIC INEQUALITIES

To solve logarithmic inequalities, we should be aware of properties of logarithm and rules followed in inequalities.

Conversion between logarithmic form to exponential form :

Note :

• When log y x > k, if y > 1, then x > yk
• When log y x > k, if 0 < y < 1, then x < yk

The problem can be solved using change base rule, the detailed example is given below.

Solve the inequality.

Problem 1 :

Solution :

Solution is (125, ∞).

Problem 2 :

Solution :

The solution is (0, 64]

Problem 3 :

Solution :

Solution is [4, ∞).

Problem 4 :

Solution :

Solution is [4, ∞).

Problem 5 :

Solution :

Solution is [64, ∞).

Problem 6 :

Solution :

Solution is [81, ∞).

Problem 7 :

Solution :

Solution is (0, 108)

Problem 8 :

 10x + 3 > 010x > -3x > -3/10 7x - 4 > 07x > 4x > 4/7
• Considering x > -7/3, all negative values should be ignored. The possible solution is (0, ∞).
• Considering x > -3/10, all negative values should be ignored. The possible solution is (0, ∞).
• Considering x > 4/7, the possible solution is (4/7, ∞).

By considering all three, the solution for the given inequality should be (4/7, ∞).

Problem 9 :

log x + log (2 - x) < 1

Solution :

log x + log(2 - x) < 1

log [x(2 - x)] < 1

Moving base to the other side of the inequality sign, we get

x(2 -x) < 101

2x - x2 < 10

-x2 + 2x - 10 < 0

x2 - 2x + 10 > 0

This quadratic inequality cannot be solved using the method of factoring.

a = 1, b = -2 and c = 10

b2 - 4ac = (-2)2 - 4(1)(10)

= 4 - 40

= -36 < 0

Domain is (0, 2). So, the solution is (0, 2).

Problem 10 :

Solution :

Domain of log2 (x2 - x - 6)

x2 - x - 6 > 0

(x - 3) (x + 2) > 0

x > 3 and x > -2

Domain of the function is (-2, ∞) u (3, ∞), solution by solving the logarithmic inequality is (-∞, 7).

By considering all these three, we get the solution as (3, 7).

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