Logarithmic Inequalities
Logarithmic inequalities are inequalities in which one (or both) sides involve a logarithm. Like exponential inequalities, they are
useful in analyzing situations involving repeated multiplication, such as in the cases of interest and exponential decay.
Contents
Introduction
Logarithmic Inequalities - Same Base
Logarithmic Inequalities - Base less than 1
Logarithmic Inequalities - Similar Base
Logarithmic Inequalities - Di erent Base
Logarithmic Inequalities - Multiple Terms
Logarithmic Inequalities - Problem Solving
See Also
Introduction
The key to working with logarithmic inequalities is the following fact:
If a
> 1 and x > y , then loga x > loga y . Otherwise, if 0 < a < 1, then loga x < loga y .
Of course, the base of a logarithm cannot be 1 or nonpositive. More importantly, the converse is true as well:
If a
> 1 and loga x > loga y , then x > y . Otherwise, if 0 < a < 1, then x < y .
In more formal terms, the logarithmic function f (x)
= loga x is monotonically increasing (increasing x always increases
f (x)) for a > 1, and monotonically decreasing (increasing x always decreases f (x)) for 0 < a < 1.
Fortunately, both of these facts are quite intuitive: when the base is greater than 1, the side with the larger argument will be the
greater one, and the opposite is true when the base is less than 1. For instance, without any knowledge of the facts formalized
above, one would intuitively expect log2 100 to be larger than log2 95.
It is also important to keep in mind the following fact:
The argument of a logarithm must be positive!
Thus, it is also necessary to take into account any inequalities resulting from the arguments being positive; for example, an
inequality involving the term log2 (2x − 3) immediately requires x
> 32 .
Logarithmic Inequalities - Same Base
When both sides of an inequality have the same base, the key facts from the introduction can be applied directly. For example,
EXAMPLE
What values of x satisfy the following inequality:
log2 (2x + 3) > log2 (3x)?
Since the base is 2, which is greater than 1, the fact that log2 (2x + 3)
2x from both sides gives 3 > x.
> log2 (3x) implies that 2x + 3 > 3x. Subtracting
Additionally, the arguments of both the logarithms must be positive, so additionally 3x
more restrictive, as x
> 0, so the final solution set is 0 < x < 3. □
> 0 and 2x + 3 > 0. The first is
TRY IT YOURSELF
True or False?
True only if n
For all positive integers n
≥ 2,
≤ 10
False always
log10 (n − 1) + log10 (n + 1) < 2 log10 n.
True only if n
≤ 100
True always
TRY IT YOURSELF
Submit your answer
1
1
<
x+1
log4 (x + 3)
log4 (
)
x+2
If the range of x that satisfy the equality above is (a, ∞), find the value of a.
This same concept can be applied to larger "stacks" of logarithms:
EXAMPLE
What values of x satisfy the following inequality:
log2 ( log3 (4x + 1)) > log2 ( log3 (2x + 3))?
Since the base is 2, which is greater than 1, the given inequality implies log3 (4x + 1)
that 4x + 1
> log3 (2x + 3). This further implies
> 2x + 3, or that 2x > 2 ⟹ x > 1. Therefore, x must be greater than 1.
In addition, the arguments of all the logarithms must be positive, but this is indeed the case when x is greater than 1.
Therefore, the solution set is x
> 1. □
TRY IT YOURSELF
Find all values of the parameter a
R:
∈ R for which the following inequality is valid for all x ∈
Submit your answer
1 + log5 (x2 + 1) ≥ log5 (ax2 + 4x + a).
If the range of values of a can be expressed in the form of (A, B], then find the value of
A + B.
Logarithmic Inequalities - Base less than 1
In the case where the base is less than 1, the previous intuition is essentially reversed: the larger side is now the one with the
smaller exponent.
EXAMPLE
What values of x satisfy the following inequality:
log 12 (3x) > log 12 (2x + 3)?
Since the base is 12 , which is less than 1, the given inequality implies 3x
Additionally, the argument of each logarithm must be positive, so 3x
the solution set is 0
< x < 3. □
Logarithmic Inequalities - Similar Base
< 2x + 3. Then x < 3.
> 0 and 2x + 3 > 0. The former is more restrictive, so
In many inequalities, the bases are di erent but can be rewri en in terms of the same base. For example,
EXAMPLE
What values of x satisfy the following inequality:
log2 (x + 1) > log4 (x2 )?
Here, the bases are di erent, but they are related by the fact that 4
= 22 . Rewriting the inequality to use 4 as a base gives
log4 ((x + 1)2 ) > log4 (x2 ),
> x2 , implying that 2x + 1 > 0 ⟹ x > − 12 . Additionally, the arguments of each logarithm must be
positive, which excludes the case x = 0. Therefore, the final solution set is x > − 12 , x 
= 0. □
so (x + 1)2
This concept can also be applied to constants, by writing c as log(10c ).
TRY IT YOURSELF
log0.2 log6 (
x2 − x
)>0
x2 + 1
(−∞, −1)
(−∞, −2)
Solve for x.
(−∞, −1) ∪ (1, ∞)
(−∞, −1] ∪ (1, ∞)
3
(−∞, − )
2
TRY IT YOURSELF
log3x+5 (9x2 + 8x + 8) > 2
Solve the inequality above for x.
4 17
x∈( , )
3 22
4
17
x ∈ ( ,− )
3
22
Try Fun with inequalities-1 and Fun with inequalities-3.
4 17
x ∈ (− , )
3 22
4
17
x ∈ (− , − )
3
22
The same concept can be applied when there are several terms:
EXAMPLE
What values of x satisfy the following inequality:
log2 (x) > log2 (3) + log4 (25) + log8 (343)?
Writing all the logarithms in base 2 gives
log2 (x) > log2 (3) + log2 (5) + log2 (7)
⇒ log2 (x) > log2 (105),
so x
> 105. All x greater than 105 satisfy the original inequality. □
Logarithmic Inequalities - Di erent Base
When the bases are di erent and not related by a common base (as in the previous section), the use of the change of base
formula becomes necessary. For example,
EXAMPLE
What values of x satisfy the following inequality:
log7 (x + 5) > log5 (x + 5)?
By change of base, the inequality gives
log(x + 5)
log(x + 5)
>
.
log 7
log 5
This is true exactly when log(x + 5) is negative, meaning that
log(x + 5) < 0 ⟹ x + 5 < 1 ⟹ x < −4,
and since x + 5 must be positive, x
> −5, and the final solution set is −5 < x < −4. □
TRY IT YOURSELF
If log0.3 (x − 1)
< log0.09 (x − 1), then x lies in the interval __________.
2<x<∞
−2 < x < −1
−∞ < x < 2
1<x<2
TRY IT YOURSELF
Submit your answer
log5n 30 5 ≥ log4n 48
Over the domain n
> 1, let M be the smallest value of n that satisfies the above
inequality.
What is M 3 ?
Logarithmic Inequalities - Multiple Terms
In the case of multiple terms, it is generally worth assigning another variable to a logarithmic term, solving the resulting
inequality, and then working with the single-term inequality. For example,
EXAMPLE
What values of x satisfy the following inequality:
2
log2 (x) + ( log2 (x)) > 6?
= log2 (x), so that y + y 2 > 6. This can be rearranged as y 2 + y − 6 = (y − 2)(y + 3) > 0, which is true when
y > 2 or y < −3.
Let y
Thus, either log2 (x)
1
.
8
> 2 ⟹ log2 (x) > log2 4 ⟹ x > 4, or log2 (x) < −3 ⟹ log2 (x) < log2 ( 18 ) ⟹ x <
The solution set is thus 0
TRY IT YOURSELF
<x<
1
and x
8
> 4, since the argument of a logarithm must be positive. □
log4 x2 + logx 16 − 5 < 0
Submit your answer
How many positive integer solutions can satisfy the inequality above?
In the case of an inequality chain, it is usually appropriate to treat each inequality separately, then combine the results. For
example,
EXAMPLE
What values of x satisfy the following inequality:
log 12 (x + 2) < −2 < log 14 (2x)?
The first inequality is log 1 (x + 2)
2
2 > 4, so x > 2.
The second inequality is −2
or 8
> x.
Thus, 2
< −2, or log 12 (x + 2) < log 12 (4). Since the base
1
2 is less than 1, this implies that x
+
< log 14 (2x), or log 14 (16) < log 14 (2x). Again, the base is less than 1, so this implies 16 > 2x
< x < 8. Since the argument of each logarithm is positive in this interval, the final solution set is 2 < x < 8. □
Logarithmic Inequalities - Problem Solving
Keep in mind that the base of a logarithm can be less than 1, related by the equality
loga b = − log a1 b,
so don't forget this case!
EXAMPLE
Solve the inequality
log x+4
(log2
2
2x − 1
) < 0.
3+x
> −4 and x =
 −2. Similarly, the argument
or x < −3. Thus the only possible values of x are x > 12 or −
Firstly, the base must be positive and not equal to 1, so immediately x
2x−1
log2 2x−1
3+x must be positive, so x+3
>0 ⟹ x>
4 < x < −3.
1
2
Now consider two cases:
Case 1. 0
<
x+4
2
< 1 ⟺ −4 < x < −2
In this case, it is necessary that log2 2x−1
3+x
So all x such that −4
> 1, or
2x−1
3+x
> 2, so
< x < −3 satisfy the original inequality.
2x−1
3+x
−2=
−7
3+x
> 0. Hence 3 + x is negative, or x < −3.
Remember that this strategy of subtracting the two sides rather than multiplying avoids the casework involved through
multiplication, as it is no longer necessary to consider the implications of the multiplied quantity being negative.
Case 2. x+4
2
> 1 ⟺ x > −2
In this case, it is necessary that log2 2x−1
3+x
2x−1
−7
< 1, or 2x−1
3+x < 2, so 3+x − 2 = 3+x < 0. This is always true as x > −2, so any
x > −2 would work. However, recall that only x > 12 or − 4 < x < −3 could possibly work, and all x such that x > 12 do.
Therefore, the solution set is
x>
1
, −4 < x < −3. □
2
A typical problem-solving strategy is using the change-of-base formula to make all the logarithms have the same base. This
makes it much easier to apply other inequalities, such as AM-GM.
EXAMPLE
Show that
logn (n + 1) > logn+1 (n + 2)
for all integers n
≥ 2.
Using change of base,
logn (n + 1) > logn+1 (n + 2) ⟺
log(n + 1)
log(n + 2)
>
log n
log(n + 1)
2
⟺ ( log(n + 1)) > log n log(n + 2).
Products of logarithms are a huge sign to use AM-GM, since the sum of logarithms is very easy to deal with.
In particular,
log n + log(n + 2)
≥
2
But log (n(n + 2))
log n log(n + 2) ⟹
log (n(n + 2))
≥
2
log n log(n + 2) .
= log (n2 + 2n) < log(n2 + 2n + 1) = log ((n + 1)2 ) = 2 log(n + 1), so
log(n + 1) >
log (n(n + 2))
≥
2
log n log(n + 2) .
Therefore,
2
( log(n + 1)) > log n log(n + 2),
proving the original inequality. □
See Also
Logarithms
Properties of Logarithms - Basic
Properties of Logarithms - Intermediate
Exponential Inequalities
Cite as: Logarithmic Inequalities. Brilliant.org. Retrieved 12:45, January 6, 2020, from h ps://brilliant.org/wiki/logarithmic-inequalities/