Logarithmic Inequalities

Logarithmic Inequalities

How to solve logarithmic inequalities: examples, and their solutions.

Example 1: Solve log7 (x + 2) ≤ 1

Solve the given inequality. log_7 (x + 2) <= 1

Before solving the inequality,
write the condition of the domain.

The domain of log7 (x + 2) is
(x + 2).

So x + 2 > 0.

So x > -2.

Logarithmic equations

Solve the given inequality.

Solve the log.
(base: 7)

Then x + 2 ≤ 71.

The base [7] satisfies
[base] > 1.

So the order of the inequality sign
does not change.

Logarithmic form

71 = 7

Move +2 to the right side.

Then x ≤ 5.

x > -2
x ≤ 5

Draw these two inequalities
on a number line.

Then the intersecting region is
-2 < x ≤ 5.

Example 2: Solve log0.1 (x - 3) > 2

Solve the given inequality. log_0.1 (x - 3) > 2

Before solving the inequality,
write the condition of the domain.

The domain of log0.1 (x - 3) is
(x - 3).

So x - 3 > 0.

So x > 3.

Solve the given inequality.

Solve the log.
(base: 0.1)

Then x - 3 < 0.12.

The base [0.1] satisfies
0 < [base] < 1.

So the order of the inequality sign
does change.

0.12 = 0.01

Move -3 to the right side.

Then x < 3.01.

x > 3
x < 3.01

Draw these two inequalities
on a number line.

Then the intersecting region is
3 < x < 3.01.

Example 3: Solve 2 log3 x ≥ log3 (x + 6) + 1

Solve the given inequality. 2 log_3 x >= log_3 (x + 6) + 1

Before solving the inequality,
write the conditions of the domains.

First, the domain of log3 x is
x.

So x > 0.

Next, the domain of log3 (x + 6) is
(x + 6).

So x + 6 > 0.

So x > -6.

x > 0
x > -6

Find the intersection of these two inequalities
by drawing the inequalities
on a number line.

Then x > 0.

Solve the given inequality.

2 log3 x = log3 x2

Logarithm of a power

+1 = +log3 3

Logarithm of the base

log3 (x + 6) + log3 3 = log3 (x + 6)⋅3

Logarithm of a product

The bases of the logs on both sides
are the same: 3.

Then x2 ≥ (x + 6)⋅3.

The base [3] satisfies
[base] > 1.

So the order of the inequality sign
does not change.

(x + 6)⋅3 = 3x + 18

Move 3x + 18 to the left side.

Then x2 - 3x - 18 ≥ 0.

Factor x2 - 3x - 18.

Factor a quadratic trinomial

Find a pair of numbers
whose product is the constant term [-18]
and whose sum is the middle term's coefficient [-3].

The constant term is (-).
So the signs of the numbers are different:
one is (+), and the other is (-).

(-18, 1) and (-9, 2)
are not the right numbers.

[-18] = -6⋅3
-6 + 3 = [-3]
So -6 and 3 are the right numbers.

Use -6 and +3
to write the factored form:
(x - 6)(x + 3) ≥ 0.

Find the zeros of (x - 6)(x + 3) ≥ 0.

Solving a quadratic equation by factoring

1) x - 6 = 0
So x = 6.

2) x + 3 = 0
So x = -3.

So the zeros are x = 6, -3.

To find the range of the inequality,
draw y = (x - 6)(x + 3)
on the x-axis.

The coefficient of x2 is (+): 1.
And its zeros are -3 and 6.

So draw a parabola
that is opened upward
and that passes through x = -3, 6 on the x-axis.

Solving quadratic inequalities

Find the range that satisfies
y = (x - 6)(x + 3) ≥ 0.

y is [greater than or equal to] 0.

So color the [upper region] of the graph,
[including the zeros].

Then x ≤ -3 or x ≥ 6.

x > 0
(from the domains condition)

x ≤ -3 or x ≥ 6
(the solution of the inequality)

Draw these two inequalities
on a number line.

Then the intersecting region is
x ≥ 6.