Logarithmic Inequality

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.

See log₇ (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.

log₇ (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 ≤ 7¹.

Logarithmic Form

7¹ = 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.

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 < 0.1².

0.1² = 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.

See log₃ x.

x is in the log.

So x > 0.

Next, see log₃ (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 log₃.

So change the right side +1
to +log₃ 3.

Logarithm of Itself

And combine the right side logs.

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

Logarithm of a Product

log₃ x ≥ log₃ (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 x² - 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

Find the zeros of (x - 6)(x + 3).

Quadratic Equation: by Factoring

Case 1) x - 6 = 0
Then x = 6.

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.