class="bi x0 y0 w1 h1"
class="bi x0 y0 w1 h1"
class="bi x1 y1 w2 h2"
class="bi x1 y1 w2 h3"
class="bi x1 y1 w3 h4"
class="bi x1 y1 w4 h3"
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

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

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
.

.
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

(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

= 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

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

+ 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
).


(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


+
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

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

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

+
• 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
⇔


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

) = −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

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

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

⇔ [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.

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

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
+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
(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

= 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
⇔



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

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

− 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

√
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


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 =

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).


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

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 =
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 −
√