Phương pháp giải phương trình bất phương trình chứa logarit và các bài toán liên quan Lý Văn Đức.

f(x) = log
a
x, 0 < a = 1 a
D = (0; +∞) I = R
0 < a = 1
f(x) = log
a
x
x > 0
f

(x) =
1
x ln a
.
f(x) = log
a
x
a > 1
ln a > 0
f

(x) =
1
x ln a

> 0, ∀x > 0.
a > 1 f(x) = log
a
x
f(1) = 0, f(a) = 1 lim
x→0
+
log
a
x = −∞; lim
x→+∞
log
a
x = +∞.
x 0 1 a +∞
y = log
a
x
−∞
0 
1 
+∞
0 < a < 1
f

(x) < 0, ∀x ∈ D
0 < a < 1 f(x) = log
a
x
x 0 1 a +∞

y = log
a
x
+∞
0
1
−∞
f(x) = log
a
x D = R
+
a > 1
0 < a < 1.
a > 0 a = 1 x
1
, x
2
∈ (0; +∞)
log
a
(x
1
x
2
) = log
a
x
1
+ log
a

x
2
log
a
x
1
x
2
= log
a
x
1
− log
a
x
2
.
a > 0 a = 1 x > 0 α
log
a
x
α
= αlog
a
x.
0 < a = 1, 0 < c = 1 x > 0
log
a
x =
log

c
x
log
c
x
.
f(x) = log
a
x (0 < a = 1)
x ∈ (0; +∞) (log
a
x)

=
1
x ln a
. u = u(x)
J ∈ R y = log
a
u(x) (0 < a = 1)
(log
a
u(x))

=
u

(x)
u(x) ln a
.

a > 0 a = 1 x
1
, x
2
∈ (0; +∞)
• a > 1 log
a
x
1
< log
a
x
2
⇔ x
1
< x
2
.
• 0 < a < 1 log
a
x
1
< log
a
x
2
⇔ x
1
> x
2

.
y = f(x) [a; b] f(a).f(b) < 0
c ∈ (a; b) f(c) = 0
y = f(x) [a; b] f(a) = A, f(b) = B
y = f(x) [a; b]
f : [a; b] → R
[a; b] (a; b) f(a) = f(b) c ∈ (a; b)
f

(c) = 0
f(x) [a; b] f(x)
[a; b]
M = m f(x) [a; b] c ∈ (a; b)
f

(c) = 0
M > m f(a) = f(b) c ∈ (a; b) f(c) = m
f(c) = M f

(c) = 0
f(x) (a; b) f(x)
(a; b) f

(x)
n − 1 (a; b)
f(x) (a; b) f

(x)
(a; b) f(x) (a; b)
f(x) (a; b) f


(x)
(a; b) f(x)
n + 1 (a; b)
f : [a; b] → R
[a; b] (a; b) ∃c ∈ (a; b) :
f

(c) =
f(b) − f(a)
b − a
.
F (x) = f(x) −
f(b) − f(a)
b − a
x.
F (x) [a; b] (a; b)
F (a) = F (b)
c ∈ (a; b) F

(c) = 0
F

(x) = f

(x) −
f(b) − f(a)
b − a
f


(c) =
f(b) − f(a)
b − a
F

(x) = 0 (a; b) F (x)
f(x) (a; b)
f

(x) > 0, ∀x ∈ (a; b) f(x) (a; b)
f

(x) < 0, ∀x ∈ (a; b) f(x) (a; b)
a
1
, a
2
, , a
n
b
1
, b
2
, , b
n
(a
1
b
1
+ a

2
b
2
+ + a
n
b
n
)
2
≤ (a
2
1
+ a
2
2
+ + a
2
n
)(b
2
1
+ b
2
2
+ + b
2
n
).
∃k a
i

= kb
i
, ∀i ∈ (1, 2, , n).
f(x) = (a
2
1
+a
2
2
+ +a
2
n
)x
2
−2(a
1
b
1
+a
2
b
2
+ +a
n
b
n
)x+(b
2
1
+b

2
2
+ +b
2
n
)
a
2
1
+ a
2
2
+ + a
2
n
= 0 ⇔ a
1
= a
2
= = a
n
= 0
a
2
1
+ a
2
2
+ + a
2

n
> 0 f(x)
f(x) = (a
1
x − b
1
)
2
+ (a
2
x − b
2
)
2
+ + (a
n
x − b
n
)
2
≥ 0, ∀x ∈ R.
∆

= (a
1
b
1
+ a
2
b

2
+ + a
n
b
n
)
2
− (a
2
1
+ a
2
2
+ + a
2
n
)(b
2
1
+ b
2
2
+ + b
2
n
) ≤ 0
⇔ (a
1
b
1

+ a
2
b
2
+ + a
n
b
n
)
2
≤ (a
2
1
+ a
2
2
+ + a
2
n
)(b
2
1
+ b
2
2
+ + b
2
n
).






a
1
x − b
1
= 0
a
2
x − b
2
= 0

a
n
x − b
n
= 0
∃k a
i
= kb
i
, ∀i ∈ (1, 2, , n).
x > −1

(1 + x)
α
≤ 1 + αx 0 ≤ α ≤ 1

(1 + x)
α
≥ 1 + αx α ≤ 0 ∨ α ≥ 1.
α = 0 α = 1
α < 0 α > 1 f(x) = (1 + x)
α
− αx − 1
x > −1
f

(x) = α(1 + x)
α−1
− α = α

(1 + x)
α−1
− 1

f

(x) = 0 ⇔ x = 0 f(x) ≥ 0, ∀x > −1 ⇔ (1 + x)
α
≥ 1 + αx,
∀x > −1.
0 < α < 1 f(x)
f(x) ≤ 0, ∀x > −1 ⇔ (1 + x)
α
≤ 1 + αx, ∀x > −1.
x x − 1


x
α
+ (1 − x)α ≤ 1 0 ≤ α ≤ 1
x
α
+ (1 − x)α ≥ 1 α ≤ 0 ∨α ≥ 1.
m > 0 a, b, c
a
m + 2
+
b
m + 1
+
c
m
= 0
ax
2
+ bx + c = 0
(0; 1)
F (x) =
a
m + 2
x
m+2
+
b
m + 1
x
m+1

+
c
m
x
m
[0; 1] (0; 1)
F

(x) = x
m−1
(ax
2
+ bx + c).
F (0) = F (1) = 0.
∃α ∈ (0; 1) F

(α) = 0
⇔ α
m−1
(aα
2
+ bα + c) = 0
⇔ aα
2
+ bα + c = 0.
ax
2
+ bx + c = 0 α ∈ (0; 1)
1
2014

< ln
2014
2013
<
1
2013
.
f(x) = ln x
f

(x) =
1
x
.
f(x) [2013; 2014] (2013; 2014)
c ∈ (2013; 2014)
f(2014) − f(2013)
2014 − 2013
= f

(c) ⇔ ln 2014 − ln 2013 =
1
c
⇔ ln
2014
2013
=
1
c
.

c ∈ (2013; 2014)
1
2014
<
1
c
<
1
2013
.
1
2014
< ln
2014
2013
<
1
2013
.
f(x)
a (a /∈ {0, 1, −1}) M ⊂ D(f)

∀x ∈ M ⇒ a
±1
x ∈ M,
f(ax) = f(x), ∀x ∈ M.
f(x) = sin(2πlog
2
x) f(x)
R

+
∀x ∈ R
+
2
±1
x ∈ R
+
f(2x) = sin(2πlog
2
(2x))
= sin(2π(1 + log
2
x))
= sin(2πlog
2
x) = f(x), ∀x ∈ R
+
.
f(5x) = f(x), ∀x > 0.
∀x ∈ R
+
⇒ 5
±
x ∈ R
+
log
5
(5x) = 1 + log
5
x ⇔ 2πlog

5
(5x) = 2π + log
5
x.
f(x) = cos[2πlog
5
x], ∀x > 0,
f(5x) = cos[2πlog
5
(5x)]
= cos[2π+2πlog
5
x]
= cos[2πlog
5
x] = f(x)
f(x) = cos(2πlog
5
x), ∀x > 0
R
+
f(x)
a (a /∈ {0, 1, −1}) M M ⊂ D(f)

∀x ∈ M ⇒ a
±1
x ∈ M,
f(ax) = −f(x), ∀x ∈ M.
f(3x) = −f(x), ∀x > 0.
∀x ∈ R

+
⇒ 3
±
x ∈ R
+
log
3
(3x) = 1 + log
3
x ⇔ πlog
3
(3x) = π + πlog
3
x
f(x) = cos[πlog
3
x],∀x > 0
f(3x) = cos[πlog
3
(3x)]
= cos[π + πlog
3
x]
= −cos[πlog
3
x] = −f(x).
f(x) = cos(πlog
3
x),
R

+
• log
a
f(x) = b ⇔

0 < a = 1
f(x) = a
b
.
• log
a
f(x) = log
a
g(x) ⇔

0 < a = 1
f(x) = g(x)
.
log
x
(x
2
+ 4x − 4) = 3

x
2
+ 4x − 4 > 0
0 < x = 1
⇔





x > −2 +
√
8
x < −2 −
√
8
0 < x = 1
⇔
√
8 − 2 < x = 1.
3 = log
x
x
3
(2.1) ⇔ log
x
(x
2
+ 4x − 4) = log
x
x
3
⇔ x
2
+ 4x − 4 = x
3
⇔ x

3
− x
2
− 4x + 4 = 0 ⇔ (x − 1)(x
2
− 4) = 0 ⇔

x = 1
x = 2
x = −2
x = −2
x = 1, x = 2
2log
4
(2x
2
− x + 2m − 4m
2
) + log
1
2
(x
2
+ mx − 2m
2
) = 0
x
1
, x
2

x
2
1
+ x
2
2
> 1
log
4
(2x
2
− x + 2m − 4m
2
) =
1
2
log
2
(2x
2
− x + 2m − 4m
2
)
log
1
2
(x
2
+ mx − 2m
2

) = −log
2
(x
2
+ mx − 2m
2
).
log
2
(2x
2
− x + 2m − 4m
2
) − log
2
(x
2
+ mx − 2m
2
) = 0
⇔ log
2
(2x
2
− x + 2m − 4m
2
) = log
2
(x
2

+ mx − 2m
2
)
⇔

x
2
+ mx − 2m
2
> 0
2x
2
− x + 2m − 4m
2
= x
2
+ mx − 2m
2
⇔

x
2
+ mx − 2m
2
> 0
x
2
− (m + 1)x + 2m − 2m
2
= 0

⇔



x
2
+ mx − 2m
2
> 0

x
1
= 2m
x
2
= 1 − m
.
x
1
, x
2
x
2
1
+ x
2
2
> 1




(2m)
2
+ m(2m) − 2m
2
> 0
(1 − m)
2
+ m(1 − m) − 2m
2
> 0
(2m)
2
+ (1 − m)
2
> 1
⇔

−1 < m < 0
2
5
< m <
1
2
.
t = log
a
x x > 0
log
k

a
x = t
k
,
log
x
a =
1
t
0 < x = 1.
a
log
b
c
= c
log
b
a
t = a
log
b
x
t = x
log
b
a
a
log
b
x

t = log
b
x
(x − 2)
log
3
[9(x−2)]
= 9(x − 2)
3
.
x − 2 > 0 ⇔ x > 2.
log
3
[(x − 2)
log
3
[9(x−2)]
] = log
3
[9(x − 2)
3
]
⇔ [log
3
[9(x − 2)]].log
3
(x − 2) = 2 + log
3
(x − 2)
3

⇔ [2 + log
3
(x − 2)].log
3
(x − 2) = 2 + 3log
3
(x − 2).
t = log
3
(x − 2).
(2 + t)t = 2 + 3t ⇔ t
2
− t − 2 = 0 ⇔

t = −1
t = 2
.
• t = −1 log
3
(x − 2) = −1 ⇔ x =
7
3
.
• t = 2 log
3
(x − 2) = 2 ⇔ x = 11.
x =
7
3
x = 11.

log
a
(ax).log
x
(ax) = log
a
2
(
1
a
), 0 < a = 1.

ax > 0
0 < x = 1
⇔ 0 < x = 1.
(log
a
a + log
a
x).(log
x
a + log
x
x) = −
1
2
log
a
a
⇔ (1 + log

a
x).(
1
log
a
x
+ 1) = −
1
2
.
t = log
a
x
(1 + t).(
1
t
+ 1) = −
1
2
⇔ 2t
2
+ 5t + 2 = 0 ⇔

t = −
1
2
t = −2
.
• t = −
1

2
log
a
x = −
1
2
⇔ x =
1
√
a
.
• t = −2 log
a
x = −2 ⇔ x =
1
a
2
.
x =
1
√
a
x =
1
a
2
.
∆
lg
2

x − lg x.log
2
(4x) + 2log
2
x = 0.
x > 0
lg
2
x − (2 + log
2
x) lg x + 2log
2
x = 0.
t = lg x
t
2
− (2 + log
2
x).t + 2log
2
x = 0
∆ = (2 + log
2
x)
2
− 8log
2
x = (2 − log
2
x)

2
.

t = 2
t = log
2
x
.
• t = 2 lg x = 2 ⇔ x = 100.
• t = log
2
x lg x = log
2
x ⇔ lg x =
lg x
lg 2
⇔ lg x = 0 ⇔ x = 1.
x = 100, x = 1.
lg
2
(x
2
+ 1) + (x
2
− 5). lg(x
2
+ 1) − 5x
2
= 0.
t = lg(x

2
+1) t ≥ 0 x
2
+1 ≥ 1 lg(x
2
+1) ≥ lg 1 = 0.
t
2
+ (x
2
− 5).t − 5x
2
= 0
∆ = (x
2
− 5)
2
+ 20x
2
= (x
2
+ 5)
2
.

t = 5
t = −x
2
.
• t = 5 lg(x

2
+ 1) = 5 ⇔ x
2
+ 1 = 10
5
⇔ x = ±
√
99999.
• t = −x
2
lg(x
2
+ 1) = −x
2
⇔

lg(x
2
+ 1) = 0
x
2
= 0
⇔ x = 0.
x = ±
√
99999 x = 0.
•
•

3 + log

2
(x
2
− 4x + 5) + 2

5 − log
2
(x
2
− 4x + 5) = 6.



x
2
− 4x + 5 > 0
3 + log
2
(x
2
− 4x + 5) ≥ 0
5 − log
2
(x
2
− 4x + 5) ≥ 0
⇔ x
2
− 4x + 5 ≤ 2
5

⇔ x
2
− 4x − 27 ≤ 0
⇔ 2 −
√
29 ≤ x ≤ 2 +
√
29.

u =

3 + log
2
(x
2
− 4x + 5)
v =

5 − log
2
(x
2
− 4x + 5)
, u, v ≥ 0

u + 2v = 6
u
2
+ v
2

= 8
⇔

u = 6 − 2v
(6 − 2v)
2
+ v
2
= 8
⇔

u = 6 − 2v
5v
2
− 24v + 28 = 0
⇔





u = 6 − 2v

v = 2
v =
14
5
⇔









u = 2
v = 2





u =
2
5
v =
14
5
.
•

u = 2
v = 2
⇔


3 + log
2
(x

2
− 4x + 5) = 2

5 − log
2
(x
2
− 4x + 5) = 2
⇔ log
2
(x
2
− 4x + 5) = 1 ⇔ x
2
− 4x + 5 = 2
⇔ x
2
− 4x + 3 = 0 ⇔

x = 1
x = 3
.
•

u =
2
5
v =
14
5

⇔


3 + log
2
(x
2
− 4x + 5) =
2
5

5 − log
2
(x
2
− 4x + 5) =
14
5
⇔ log
2
(x
2
− 4x + 5) = −
71
25
⇔ x
2
− 4x + 5 = 2
−
71

25
⇔ x
2
− 4x + 5 − 2
−
71
25
= 0.
log
2
(x −

x
2
− 1) + 3log
2
(x +

x
2
− 1) = 2.



x
2
− 1 ≥ 0
x −
√
x

2
− 1 > 0
x +
√
x
2
− 1 > 0
⇔ x ≥ 1.

u = log
2
(x −
√
x
2
− 1)
v = log
2
(x +
√
x
2
− 1)
.
u + v = log
2
(x −

x
2

− 1) + log
2
(x +

x
2
− 1)
= log
2

(x −

x
2
− 1).(x +

x
2
− 1)

= log
2
1 = 0.

u + v = 0
u + 3v = 2
⇔

u = −v
2v = 2

⇔

u = −1
v = 1
⇔

log
2
(x −
√
x
2
− 1) = −1
log
2
(x +
√
x
2
− 1) = 1
⇔

x −
√
x
2
− 1 =
1
2
x +

√
x
2
− 1 = 2
⇔ x =
5
4
.
x =
5
4
.
f [x, ϕ(x)] = 0.
u = ϕ(x)

u = ϕ(x)
f(x, u) = 0
.
log
2
2
x +

log
2
x + 1 = 1.
u = log
2
x.
u

2
+
√
u + 1 = 1

u + 1 ≥ 0
1 − u
2
≥ 0
⇔ −1 ≤ u ≤ 1.
v =
√
u + 1 0 ≤ v ≤
√
2 v
2
= u + 1.

u
2
= 1 − v
v
2
= u + 1
⇒ u
2
− v
2
= −(u + v)
⇔ (u + v)(u −v + 1) = 0 ⇔


u + v = 0
u − v + 1 = 0
.
• v = −u
u
2
− u − 1 = 0 ⇔




u =
1 −
√
5
2
u =
1 +
√
5
2
.
u =
1 +
√
5
2
−1 ≤ u ≤ 1
u =

1 −
√
5
2
⇔ log
2
x =
1 −
√
5
2
⇔ x = 2
1 −
√
5
2
.
• u − v + 1 = 0
u
2
+ u = 0 ⇔

u = 0
u = −1
⇔

log
2
x = 0
log

2
x = −1
⇔

x = 1
x =
1
2
.
x = 2
1 −
√
5
2
, x = 1 x =
1
2
7
x−1
= 6log
7
(6x − 5) + 1.
6x − 5 > 0 ⇔ x >
5
6
.
y − 1 = log
7
(6x − 5).


7
x−1
= 6(y − 1) + 1
y − 1 = log
7
(6x − 5)
⇔

7
x−1
= 6y − 5
7
y−1
= 6x − 5
.
7
x−1
+ 6x = 7
y−1
+ 6y.
f(t) = 7
t−1
+ 6t
f(x) = f(y) ⇔ x = y.
7
y−1
= 6x − 5 7
x−1
− 6x + 5 = 0.
g(x) = 7

x−1
− 6x + 5
• D = (
5
6
; +∞).
•
g

(x) = 7
x−1
. ln 7 − 6
g

(x) = 7
x−1
.ln
2
7 > 0, ∀x ∈ D
g

(x)
g(x) = 0
g(1) = g(2) = 0
x = 1 x = 2
s
ax+b
= clog
s
(dx + e) + αx + β

d = ac + α e = bc + β

0 < s = 1
dx + e > 0
.
ay + b = log
s
(dx + e).
lg
4
x + lg
3
x − 2lg
2
x − 9 lg x − 9 = 0.
x > 0
t = lg x
t
4
+ t
3
− 2t
2
− 9t − 9 = 0 ⇔ 3
2
+ 3t.3 − t
4
− t
3
+ 2t

2
= 0.
u = 3
u
2
+ 3t.u − t
4
− t
3
+ 2t
2
= 0.
∆ = 9t
2
+ 4(t
4
+ t
3
− 2t
2
) = (2t
2
+ t)
2



u =
−3t − (2t
2

+ t)
2
u =
−3t + (2t
2
+ t)
2
⇔

u = −t
2
− 2t
u = t
2
− t
⇔

3 = −t
2
− 2t
3 = t
2
− t
⇔ t
2
− t − 3 = 0 ⇔ t =
1 ±
√
13
2

.
• t =
1 +
√
13
2
lg x =
1 +
√
13
2
⇔ x = 10
1 +
√
13
2
.
• t =
1 −
√
13
2
lg x =
1 −
√
13
2
⇔ x = 10
1 −
√

13
2
.
x = 10
1 ±
√
13
2
lg
4
x + (2m − 1)lg
3
x + m(m − 2)lg
2
x − (m
2
− m + 1) lg x − m + 1 = 0.