# Logarithmic Inequality

How to solve the logarithmic inequality: formula, 3 examples, and their solutions.

## Formula

For log_{a} x,

x > 0

0 < a < 1, a > 1.

Just like solving a logarithmic equation,

when solving a logarithmic inequality,

the solution should satisfy these conditions.

## Examplelog_{7} (x + 2) ≤ 1

See log_{7} (x + 2).

(x + 2) is in the log.

So x + 2 > 0.

Move +2 to the left side.

Then x > -2.

So x should satisfy this condition:

x > -2.

Next, solve the inequality.

log_{7} (x + 2) ≤ 1

The exponent is the right side 1.

The base is 7.

7 is greater than 1.

(not between 0 and 1)

So the order of the inequality sign doesn't change.

Then x + 2 ≤ 7^{1}.

Logarithmic Form

7^{1} = 7

Move +2 to the right side.

Then x ≤ 5.

Then, graph the found inequalities.

The condition you found first is

x > -2.

And you got x ≤ 5.

Graph these two inequalities

on a number line.

And color the intersecting region.

The colored region is

-2 < x ≤ 5.

So

-2 < x ≤ 5

is the answer.

## Examplelog_{0.1} (x - 3) > 2

See log_{0.1} (x - 3).

(x - 3) is in the log.

So x - 3 > 0.

Move -3 to the left side.

Then x > 3.

So x should satisfy this condition:

x > 3.

Next, solve the inequality.

log_{0.1} (x - 3) > 2

The exponent is the right side 2.

The base is 0.01.

0.01 is between 0 and 1.

(not greater than 1)

So the order of the inequality sign does change.

Then x - 3 < 0.1^{2}.

0.1^{2} = 0.01

Move -3 to the right side.

Then x < 3.01.

Then, graph the found inequalities.

The condition you found first is

x > 3.

And you got x < 3.01.

Graph these two inequalities

on a number line.

And color the intersecting region.

The colored region is

3 < x < 3.01.

So

3 < x < 3.01

is the answer.

## Example2 log_{3} x ≥ log_{3} (x + 6) + 1

See log_{3} x.

x is in the log.

So x > 0.

Next, see log_{3} (x + 6).

(x + 6) is in the log.

So x + 6 > 0.

Move +6 to the right side.

Then x > -6.

x > 0

x > -6

Draw these two inequalities

on a number line.

Color the intersecting region.

The colored region is x > 0.

So x should be in this colored region.

Next, solve the inequality.

The log terms have the same log_{3}.

So change the right side +1

to +log_{3} 3.

Logarithm of Itself

And combine the right side logs.

log_{3} (x + 6) + log_{3} 3 = log_{3} (x + 6)⋅3

Logarithm of a Product

log_{3} x ≥ log_{3} (x + 6)⋅3

The base is 3.

3 is greater than 1.

(not between 0 and 1)

So the order of the inequality sign doesn't change.

Then x ≥ (x + 6)⋅3.

(x + 6)⋅3 = 3x + 18

Multiply a Monomial and a Polynomial

Move 3x + 18 to the left side.

Factor the left side x^{2} - 3x - 18.

Find a pair of numbers

whose product is the constant term -18

and whose sum is the coefficient of the middle term -3.

-6⋅3 = -18

-6 + 3 = -3

So (x - 6)(x + 3) ≥ 0.

Factor a Quadratic Trinomial

Case 2) x + 3 = 0

Then x = -3.

Case 1: x = 6

Case 2: x = -3

So the zeros are x = -3, 6.

Draw y = (x - 6)(x + 3)

on the x-axis.

First point the zeros x = -3 and 6.

And draw a parabola

that passes through x = -3 and 6.

Quadratic Function: Find Zeros

See (x - 6)(x + 3) ≥ 0.

The left side is greater than or equal to 0.

So color the region

where the graph is above the x-axis (y = 0).

The inequality sign includes equal to [=].

So draw full circles on x = -3, 6.

Quadratic Inequality

The colored region is

x ≤ -3 or x ≥ 6.

Then, graph the found inequalities.

The condition you found first is

x > 0.

And you got

x ≤ -3 or x ≥ 6.

Graph these two inequalities

on a number line.

And color the intersecting region.

The colored region is

x ≥ 6.

So

x ≥ 6

is the answer.