1
Logarifm
mavzusi
yechimlari
2
1.14.Logarifm.
1.14.1. Logarifmik funksiya va uning xossalari.
1(96-6-52)
)
log3 (2
x
y
;
; 2
2;
0;
2
x
x
x
(A)
2(97-1-63)
;)
log (3
x
y
x
3;1
1;0
;
1
0
3
;
1
0
0
3
x
x
x
x
x
x
x
(C)
3(97-6-64)
;)
log (6
( )
x
x
f
x
6;1
1;0
;
1
0
6
;
1
0
0
6
x
x
x
x
x
x
x
(D)
4(97-8-52)
4 ;)
1
(
log
1
x
y
x
;2
2;1
;
2
1
4
1
;
1
1
0
1
0
4
1
x
x
x
x
x
x
x
(B)
5(97-9-75)
)1
5
lg(
2
x
nx
y
funksiya berilgan
oraliqda aniqlangan bo’lishi uchun 4
1 va 1
0
1
5
2
nx x
tenglamaning ildizlari bo’lishi
kerak. 4
1 va funksiyning aniqlanish sohasiga
kiritilgan. Bunday bo’lishi mumkin emas. (D)
6(97-12-52)
)
log 3 (6
x
y
x
;
6;1
1;0
;
1
0
6
;
1
0
0
6
3
3
x
x
x
x
x
x
x
(E)
7(98-7-42)
x
x
y
5
1
5
log
log
I va IV choraklardan o’tadi. (D)
8(98-5-15)
5
3
1
;
;
y
y
y
juft funksiya. (D)
9(98-12-42)
x
y
log3
;
I va IV choraklardan o’tadi. (A)
10(99-2-36)
;)1
lg(
8
( )
x
x
f x
8;2
2;1
;
2
1
8
;
0
)1
lg(
0
1
0
8
x
x
x
x
x
x
x
. Bu
oraliqqa tegishli 6 ta butun son bor. (D)
11(99-3-26)
2 ;
)
1
ln(
1
x
x
y
1;0
;2 0
;
2
0
1
;
0
2
0
)
1
ln(
0
1
x
x
x
x
x
x
x
(D)
12(99-5-39)
x
x
x
f
1
2
8
log 64
( )
;
;
8
; 8
8
0; 64
8
64
1
2
1
1
x
x
x
x
x
x
1
;
;1
;
1
2
x
x
x
x
(B)
3
13(99-6-29)
;
log
3
log
3
3
x
x x
;3
;
0
0
3)
(
x
x
x x
(A)
14(97-7-15)
)1
2
lg(
2
x
kx
y
funkiya faqat x=1
nuqtada aniqlanmagan bo’lishi uchun
2
2
)1
(
1
2
x
x
kx
bo’lishi yoki k=1 bo’lishi
kerak. (E)
15(99-8-34)
y
;
log
1
1
2 x
x
1;0
0;
1
;
0
0
1
;
0
0
1
1
x
x
x
x
x
x
x
(A)
16(99-8-36)
27 ;)
)3
log ((
36
6
log
)
(
2
3
2
3
x
x
x
f x
3
log 27
)
(
3
min
f x
(D)
17(99-9-50)
;
9
2
4
ln
2
x
x
x
y
9;8
;0)
(
;
9
0
8)
(
;
0
9
0
2
4
2
x
x
x
x
x
x
x
;
(E)
18(00-9-44)
3
2
5
3
log 81
)
(
x
x
f x
;
;
3
; 3
3
0; 81
3
81
2 3
4
2 3
2 3
x
x
x
x
x
x
0;
)1
3)(
0 ; (
3
4
3;
4
2
2
x
x
x
x
x
x
x ;3 1
(D)
19(96-3-90)
3;
log
log 3
5;
log
2
1
4
1
2
1
b
a
3;
log
2
1
c
3;
3
5
b
c
a
(E)
20(96-9-25)
;1
log 3
;1
3
log
4
1
3
1
b
a
;1
3
log
2
1
c
b
a
c
(A)
21(96-12-90)
3
2
log 6
3 ;
2
4
log
8
1
8
1
b
a
3 ;
2
log 4
6
log
6
1
8
1
c
a
c
b
(B)
22(96-13-31)
;
log 6
4;
log
5
1
5
1
a
b
a
;
4
log
6
1
a
c
c
a
b
(E)
23(98-9-32) q va l musbat. (C)
24(02-2-20)
1
2
log
2
1
(A)
25(99-9-47)
m
n
va
p
1
1
0
bo’lsa,
0
;0 log
log
m
n
p
p
yoki
0
log
log
m
n
p
p
bo’ladi. (B)
26(0-3-43) d val musbat (E)
27(99-2-30) Hech qaysisi to’g’ri emas. (E)
28(01-1-29)
;
1
5
5)
6
ln(
2
2
x
x
x
y
;
1
5
0
5
6
;
0
1
5
0
5
6
2
2
2
2
x
x
x
x
x
x
2
0
)1
5)(
(
x
x
x
;
( ;2 5)
;1( 2)
x
(B)
29(01-3-21)
;)
(6
log
2
10
x
x
y
0;
2)
3)(
0; (
6
0;
6
2
2
x
x
x
x
x
x
3
2
x
. -1,0,1,2 butun sonlar ularning
yig’indisi 2 ga teng. (C)
30(01-6-19)
0
; 9
9
10
2
2
2
lg 9
x
x
y
x
;
I va II chorakdan o’tadi.
4
31(01-6-40)
;
2
6
log
)
(
1
2
x
x
x
f
x
6;0
0;2
;
0
2
6
;
1
1
0
2
0
6
2
x
x
x
x
x
x
x
(С)
32(01-7-44)
;
)1
lg(
9
)
(
2
x
x
x
f x
;
0
1
0
3)
3)(
(
;
0
0
1
0
9
2
x
x
x
x
x
x
x
3;1
x
(E)
33(01-9-46)
;
9
25
30
13
log
2
2
x
x
x
y
0;
9
25
30)
13
0; (
9
25
30
13
2
2
2
2
x
x
x
x
x
x
0;
3
5
3
5
15)
2)(
(
x
x
x
x
3;15
5
3
5
;2
x
. Bu oraliqqa 13 ta natural
son tegishli (A)
34(01-9-47) )
;
3
2
15
2
log
2
15
x
x
x
y
0;
5,1
15)
2
0; (
3
2
15
2
2
2
x
x
x
x
x
x
0;
5,1
)5
3)(
(
x
x
x
;5
5,1
;3
x
.
Bu oraliqqa tegishli eng kichik butun son -2.
log 7
2)
(
15
f
(E)
35(98-7-21)
2 ;)
lg(
9)
lg(
3)
lg(
x
x
x
;3
;
2
9
3
;
0
2
0
9
0
3
x
x
x
x
x
x
x
(D)
36(02-3-42)
lg 3 ;
5
5lg 3 ;
x
y
x
y
10 ;
3
5
y
x
5
5
3 10
10 ;
3
x
y
y
x
(A)
37(02-3-43)
;
6
1
)1
3
lg(
x2
x
x
y
;
0
)2
3)(
(
3
1
;
0
6
1
3
;
0
6
0
1
3
2
2
x
x
x
x
x
x
x
x
x
3;3
1
x
(A)
38(02-4-39) Funksiyaning OY o’qi bilan kesishish
nuqtasida x=0 bo’ladi.
2
4
10
(0)
lg(0 4)
y
(E)
39(02-7-20)
1
2
1
3
lg x
x
y
;
;
2
0
2
1
2
;
0
2
0
1
2
1
3
x
x
x
x
x
x
2 ;
1
; 2
;
2
0
2)
2 (
1
x
x
x
x
(A)
40(02-9-29)
;)
log (3
2
2
1
x
y
;
3
2
)
(3
log
;
0
3
0
)
log (3
2
2
1
2
1
x
x
x
x
3;1
;
3
1
;
3
4
3
x
x
x
x
x
(B)
41(02-12-51)
4 ;)
log (
4
)
(
2
2
x
x
f x
;
4
0
2)
2)(
(
;
0
4
0
4
2
x
x
x
x
x
;2
;4 2
x
(E)
42(03-4-40)
x
x
x
f
x
2
1
log
)
(
2
;
2;1
;
2
1
0
1
;
0
2
1
0
0
1
2
2
x
x
x
x
x
x
x
x
x
(D)
43(03-4-41)
4;
)1
log (
5)
2
log (
)
(
2
2
2
2
x
x
x
f x
5
2
log 4
)
(
2
min
f x
;
f (x) ;2
(E)
44(03-5-63)
log 25;
2log 5
2
2
a
27,04;
log
26
5
4log
23;
log
23
1
3log
2
4
1
2
8
1
c
b
c
a
b
(A)
4503-6-43)
;)1
lg(
1
|
|
8
2
x
x
y
0
)1
1)(
(
0
| 8
|
;
0
)1
1)(
(
0
| 0
|
8
;
0
1
0
1
|
|
8
2
x
x
x
x
x
x
x
x
x
x
;
8;1
;8 1
;
0
)1
1)(
(
0
8
8
x
x
x
x
x
(D)
46(03-10-38)
;
1
)
ln(7
2
x
x
y
;
1
0
7)
7)(
(
;
0
1
0
7
2
x
x
x
x
x
7;
;1
7; 1
x
Bu oraliqqa tegishli butun
sonlar -2,0,1,2. Bularning yig’indisi 1 (B)
1.14.2. Logarifmik ifodalarni shakl
almashtirish.
1(96-3-89)
3
2
2
2
2
2 3
log
4
2 3
log
4
1
16 3 4
log
4
316
log
1
(E)
2(96-6-53)
2
1
log 2
8
log 1
16
log
4
4
4
Q
<2 ; (D)
3(96-9-31)
9
7
7
7
log7 9
9 7
log
3
3
1
9 7
log
3
3
(B)
4(96-12-89)
4
3
3
3
log3 4
4 3
log
2
2
1
4 3
log
2
(C)
5(96-13-30)
4
5
5
5
log5 4
4 5
log
3
3
1
4 5
log
3
3
(D)
6(97-2-53)
4;
2
6
log 4
2log 8
2
2
m
log 400 log 25
2log 5
400
log
2
2
2
2
n
4
log2 16
;
4;
3 1
log 5
log 125
5
5
p
;1
ln
ln12
ln12
e
e
q
q
p
m
n
(D)
7(975-37)
;1
log 2
lg100
log
2
2
(A)
8(97-8-53)
2
0
log 25
log 25
log 25
2log 5
2
2
2
2
(C)
9(97-9-37)
1
log 5
ln
log
5
5
5
e
(E)
10(97-12-52)
2log 5 log 30
2
2
1
30
log 25
log 30
25
log
2
2
2
; (D)
11(98-4-15)
,0 05
20
1
20
20
5
20
5
5
lg
20
lg
5 1
lg
20
lg
(D)
12(98-9-73)
3
log 729
9
log
729
log
9
2
2
(B)
13(98-11-46)
2log 128
128
log
8
2 2
3 ;
4 2
3
14
2
2log
7
23
(A)
14(98-12-74)
3
log
4
log
3
log
12
log
108
3
36
3
=
log 4 log 108
log 12 log 36
3
3
3
3
=
log 4 log (27 4)
log (4 3) log (9 4)
3
3
3
3
=
log 4)
log 4 (3
log 4)
)1 (2
4
(log
3
3
3
3
=
log 4
3log 4
log 4
2
log 4
2log 4
2
3
3
3
2
3
3
=2 (B)
15(99-2-31)
10
10
10
100
lg27 2lg3
2 lg27 lg3
1
30
3 10
10
10
10
10
lg3
lg27 lg9
(C)
16(99-3-15)
3
3
3
5
9
5
9
5
2
1
2
1
5 3
log
1
1
log5 3
2
4
1
2 1
5,0
log
1
4 3
5
log
(B)
17(99-6-13)
7 log 3
log 17 log
7
17
9
9
log
17
log
3
3
6
14
7
2
1
9
log
3
log
7
log
3
log
17
log
7
log
3
3
3
3
3
3
(A)
18(99-8-30)
1
2
1
lg ,0 026
lg28
(C)
19(00-1-35)
9
7
2
9
25
10
100
2
15
lg25
2lg5 lg15
(D)
20(00-1-39)
A)
1
log 2
log 9
log 18
2
2
2
;
B)
8,4
3
3log 2
1
3 log 2
3log 6
3
3
3
;
C)
2
lg100
lg4
lg25
;
D)
4
log 13
169
log
4
13
2
13
;
E)
3
1
4
log
64
log
4
log
64
8
8
; Javob: (B)
21(00-3-34)
8
4
7
343
2
3
2
3
log7 4
log49 4
(A)
22(00-7-32) n=
3
1
2
log 2
4
log
2
1
2
1
;
m=
;1
log 3
log 3
15
log
3
1
3
1
3
1
p=
2;
ln
2
e
-3<-2<-1 ; n<m<p (A)
23(00-10-42)
2log 512
512
log
8
2 2
=
;6
3
18
2
2log
9
23
(B)
24(96-9-84)
4 log 5 log 6 log 7 log 8 log 9
log
8
7
6
5
4
3
8
log
9
log
7
log
8
log
6
log
7
log
5
log
6
log
4
log
5
log
3
log
4
log
3
3
3
3
3
3
3
3
3
3
3
3
2
3
log
9
log
3
3
(D)
25(00-5-66)
2 log 3 log 4 log 5 log 6 log 7
log
8
7
6
5
4
3
8
log
7
log
7
log
6
log
6
log
5
log
5
log
4
log
4
log
3
log
3
log
2
log
2
2
2
2
2
2
2
2
2
2
2
2
3
1
8
log
2
log
2
2
(B)
26(00-7-31)
8
4
2
81
1
2log 3 log 2 log
3
3
2
(D)
27(00-8-43)
8 7
log
1
6 5
log
1
49
25
=
2log7 8
2log5 6
7
5
10
8
6
2
2
(A)
28(00-8-46)
log6 36
1 lg2
log6 5
3
10
36
=
21
9
5
25
3
10
10
6
2
2
lg
2log6 5
(A)
29(98-1-33)
2log 7
14
log
2log 7
log 14 log 7
14
log
2
2
2
2
2
2
2
2
14 2log 7
log
2log 7
log 14 log 7
14 2log 14 log 7
log
2
2
2
2
2
2
2
2
2
2
14 2log 7
log
)7
log 7(log 14 2log
)7
14(log 14 2log
log
2
2
2
2
2
2
2
2
1
7
14
log
14 2log 7
log
)7
14 2log 7)(log 14 log
(log
2
2
2
2
2
2
2
(D)
30(98-2-36)
log 81
16
1
log
log 9 log 48
27
log
3
3
3
3
3
5
4
1
4
16
48 1
3 log
9
log
3
3
(D)
31(98-8-33)
log 18
log 2
2
log 2 log 18
2log 18
log 14
2
3
3
3
3
2
3
2
3
2log 18
log 2
2
log 2 2log 2 log 18 log 2 log 18 2log 18
2
3
3
2
3
3
3
3
3
2
3
2log 18
log 2
2
log 18(log 2 log 18)
log 2(log 2 log 18)
2
3
3
3
3
3
3
3
3
2
18
2
log
log 2 log 18
2
2 log 18)(2log 2 log 18)
(log
3
3
3
3
3
3
3
(C)
32(99-1-28)
2 1
6
log
2 1
6 log6
log
1
2
1
2 1
2 1 log6
log
)1
1)( 2
2
(
1
2
7
1
2
log
1
2
6
2 1
log 2 1 log6
(A)
33(99-4-55)
log 3
15
log
2log 3
2log 15
log 3
15
log
5
5
5
5
2
5
2
5
15 log 3
log
)3
2(log 15 log
)3
15 log 3)(log 15 log
(log
5
5
5
5
5
5
5
5
)3
15 log
(log
)2
15 log 3)(log 15 log 3
(log
5
5
5
5
5
5
3
1 2
2
3
15
log5
(C)
34(96-1-33)
log 5;
log 5;
5; log 2
2
2
2
2
n
n
n
1
1
log 5
1
1
(2 5)
log
1
2
lg
2
2
n
(E)
35(96-10-36)
2 log 5;
3
5
log
125
log
2
3
22
4
a
3 ;
2
5
log2
a
3
2
18
3
2
1
6
log 5
1
6
10
log
log 64
64
lg
2
2
2
a
a
(C)
36(96-9-28)
log 25
2
log
log 5
8
log
50
log
log 40
40
log
2
2
2
2
2
2
50
a
;
log 5;
3
2 log 5
2log 5;
1
log 5
3
2
2
2
2
a
a
a
1 ;
2
3
; log 5
3
1
5 2
log
2
2
a
a
a
a
3 ;
2
1
log5 2
a
a
(E)
37(96-3-86)
log 49
2
log
log 7
8
log
98
log
log 56
56
log
7
7
7
7
7
7
98
a
;
;1
3log 2
2
log 2
2 ;
2
log
1
log 2
3
7
7
7
7
a
a
a
;
3
1
2
2 ; log 2
1
3
2
log
7
7
a
a
a
a
(B)
38(96-12-86)
log 49
3
log
log 7
9
log
147
log
log 63
63
log
7
7
7
7
7
7
147
a
;
;1
2log 3
2
log 3
2 ;
3
log
log 3 1
2
7
7
7
7
a
a
a
2 ;
2
1
2 ; log 3
1
2
3
log
7
7
a
a
a
a
(A)
39(96-13-27)
log 25
3
log
log 5
9
log
75
log
log 45
45
log
5
5
5
5
5
5
75
a
;
;1
2log 3
2
log 3
2 ;
3
log
log 3 1
2
5
5
5
5
a
a
a
2 ;
2
1
2 ; log 3
1
2
3
log
5
5
a
a
a
a
(A)
40(98-2-33)
3,0 ;
4,2 ; log 2
8log 2
log 2
256
log
8
a
a
a
a
3,0
log 2
log 2
log 4
a
a
a
(E)
41(98-3-30)
log 9
4
log
log 27
4
log
36
log
log 108
108
log
2
2
2
2
2
2
36
a
;
3log 3;
2
2 log 3
2log 3; 2
2
3log 3
2
2
2
2
2
a
a
a
3 ;
2
)
2 1(
2 ; log 2
2
3
3 2
log
7
2
a
a
a
a
(C)
42(98-10-77)
log 3
2
log
log 27
4
log
6
log
log 108
108
log
2
2
2
2
2
2
6
a
;
3log 3;
2
log 3
log 3 ;
1
3log 3
2
2
2
2
2
a
a
a
;
3
2
; log 3
2
3
3
log
2
2
a
a
a
a
(D)
43(00-1-88)
;
log 3
4
log
1
12
log
1
2
log
2
2
2
12
a
2 ;
1
log 3; log 3
2
1
2
2
a
a
a
8
a
2a
1
1
4
log 3
2
log
4
6
log
log 16
16
log
2
2
2
2
6
a
a
1
4
(C)
44(00-1-41)
a
a
2
2
8
2
2
log log
)
log (log
a
a
2
2
2
2
log log
)
log (8log
=
3
log log
log log
8
log
2
2
2
2
2
a
a
(C)
45(00-6-32)
5,0
log
3
5,0
log
log 27
27
log
3
3
3
5,0
a
;
3 ;
5,0
log3
a
5,0
3 log
1
3 log 3
1
5,1
6 log
2
5,1
log
3
3
3
6
3
1
3
1
1
3
1
3
3
1
3
1
a
a
a
(A)
46(00-10-31)
a log2 3;
log 4
3 log 3
1
4
3 log 3
1
,0 75
log
2
2
2
8
2
3
1
2
3 log 3
1
2
a
(C)
47(00-10-66)
3 ;
3 ;
;
27
log
3
3
b
b
a
a
a
b
b
b
a
a
b
1
3
3
1
3 log 3
1
6 log
2
log
3
3
3
6
3
(A)
48(97-4-33)
;
; lg7
lg2
b
a
lg5
7
lg
5
lg
5
lg
lg7
1
1
log 7
1
1
35
log
1
5
log
5
5
35
b
a
a
1
1
lg7
lg2
1
lg2
1
(B)
49(97-9-73)
;
; log 5
20
log
3
3
b
a
log 4;
1
log 4
log 5
log 20
5
log
20
log
5
5
5
5
3
3
b
a
;
; log 5
4
log
4
5
b
a
b
b
b
a
1 3log 5
log 125
log 4
log 500
4
4
4
4
b
a
b
a
b
a
b
2
3
1
(C)
50(98-11-44)
;
; log 10
2
log
2
7
b
a
log 10
2 log 392
1
10
2 log 392
1
39 2,
log
2
2
2
4
3 log 10
log 7
2
2
1
49 log 8 log 10
log
2
1
2
2
2
2
2
2
2
3
1
3
2
2
1
b
a
b
a
(B)
51(99-7-30)
;
; log 5
3
log
2
2
b
a
log 5
9
log
log 5
27
log
45
log
log 135
135
log
2
2
2
2
2
2
45
b
a
b
a
2
3
log 5
log 3
2
log 5
log 3
3
2
2
2
2
(A)
52(99-10-35)
;9
2;
log
;4
2;
log
3
2
b
b
a
a
2
log 36
log (9 4)
)
log (
6
6
6
ab
(E)
53(00-3-33)
6
3; log
2; log
log
x
x
x
c
b
a
;
c
b
a
abc
x
x
x
x
x
abc
log
log
log
1
log
1
log
1
6
1
3
1
2
1
1
(C)
54(00-3-35)
;
; log 5
7
log
14
14
b
a
log 5
7
log
log 7
2
(7 5)
log
7
196
log
35
log
28
log
28
log
14
14
14
14
14
14
14
35
b
a
a
2
(A)
55(00-8-38)
;
; lg3
lg5
b
a
b
a
1
3
3
lg3
1
3lg5
3
)3
10
lg(
125
1000
lg
30
lg
lg8
log30 8
(A)
56(00-10-39)
;
; log 10
2
log
2
7
b
a
log 10
(49 16)
log
2
1
10
784
log
2
1
78 4,
log
2
2
2
4
9
2
2
1
4
2 2log 7
1
2
b
a
b
(D)
57(01-2-29)
2
1
2
1
4
4
2
2
log
log
b
a
a
b
2
1
2
1
2
2
4
4
2
log
2log
log
log
b
a
b
a
a
b
a
b
2
1
2
1
2
2
2
2
log
log
b
a
a
b
2
1
2
2
2
log
log
b
a
a
b
2
1
2
2
log
2log
log
log
b
a
b
a
a
b
a
b
;|
log
| log
log
log
2
1
2
b
a
b
a
a
b
a
b
b a 1
bo’lgani uchun
a
b
b
a
b
a
a
b
log
log
|
log
| log
(A)
58(01-3-14)
4
8
2 2
4
4
6
4
3
2
3
2 2
log2
(C)
59(01-3-28)
;
7
0;
0;
2
2
ab
b
a
b
a
;
3
;
9
)
(
2
ab
b
ab a
b
a
1
)
lg(
)
lg(
)
lg(
lg
2
lg
lg
3
lg
2
ab
ab
ab
ab
b
a
b
a
(A)
60(01-5-16)
4
1
4
49
5
1
7
49
5
49
log5 4
7 2 2
log
log5 4
1 log7 2
12 5,
4
50
(A)
61(01-6-36)
log 3
log 15
log 20
2log 12
2
2
2
2
3
15
log 144 20
log 3
log 15
log 20
log 144
2
2
2
2
2
6
log2 64
(E)
62(01-6-37)
log 9 log 5
lg8 log 10
3
5
2
5
2
3
5 log 5
log
log 9
10 log 10
log
8
log
3
3
3
2
2
2
(D)
63(01-7-29)
8,0 1 9log3 8 log65 5
log65 5
3log3 8 2
1
8,0
=
4
5
8,0
65
8,0
log65 5
(C)
64(01-7-24)
;
; lg7
lg2
b
a
lg2
1
1
lg2
lg7
2
5
lg
1
98
lg
10
log 98
8,9
log
5
5
=
a
b
a
1
1
2
(B)
65(01-8-31)
;
3
3
; log
3
; 3log
27
log
2,0
2,0
2,0
a
a
a
2,0 )
log
3 (2
1
2,0 )
6 log (9
2
8,1
log
3
3
6
3
a
a
1
3
2
3
3 2
1
(E)
66(01-9-10)
2
3
2
lg
3
2
lg
3
2
lg
4 3
7
lg
2
(A)
67(01-9-17)
16
9
2 lg
1
lg
8
9lg
lg
)
(
lg
)
(
lg
3
2
2
3
3
2
x
x
x
x
x
x
(A)
68(01-9-26) 4+
3 ;
16
4
1
1
4
4
1
1
16
3
3
16
1
5
1
5
1
2,0
3
16
5
log
3
16
5
log
4
5 4 1 1
log
(A)
69(01-10-16)
6 ;
1
2
1
1
4
1
32
1
16
1
8
1
4
1
2
6
1
8
log
6
1
log8
2
32
1
16
1
8
1
4
1
2
2
log
8
1
8
1
125
,0
36
6
1
1
2
(C)
10
70(01-10-33)
3
4
2
5
4
2
8
3
4
5
2
3
log 5
4
log
log 3
2
log
125
log
16
1
log
81
1
256 log
log
3
51
3
16
3 log 4 log 5
2
8 log 2 log 3
4
4
5
2
3
(С)
71(01-11-25)
2 log 243 log 5 log 4
log
3
2
4
5
4log 3 log 4
log 4
2 log 5 log 3
log
3
4
3
5
4
2
5
=5 (С)
72(01-11-26)
13
1300
lg
lg )5
(lg2
3
1300 lg13
lg
3lg2 3lg5
5,1
2
3
100
lg
3lg10
(E)
73(01-11-27)
;
; log 4
log 4
5
3
b
a
log 5
2log 3
log 5
log 9
log 45
4
4
4
4
4
ab
a
b
b
a
b
a
2
1
2
1
2 1
(E)
74(01-11-55)
t
3
3
3
2
245
2
83
log
;
x
3
3
3
2
245
2
83
log
bo’lsin.
t x
3
3
3
2
245
2
83
log
+
3
3
3
2
245
2
83
log
3
3
3
3
3
2
245
2
83
2
245
2
83
log
3
3
3
3
3
3
241
81
log
2
245
2
83
log
3;
log 3
3
log ( 3
3
3
3
5
4
3
t
x
x
t
3
3;
(D)
75(02-2-21)
;
lg5
c
1
2
1
2lg5
lg10
lg25
lg250
c
(A)
76(02-2-53)
5
log
150
log
5
log
30
log
6
5
30
5
log 30)
log 6(log 5
30
log
5
5
5
2
5
log 6 log 30
log 6
30
log
5
5
5
2
5
log 6
log 6)
30(log 30
log
5
5
5
5
1
log 5
log 6
30
log
5
5
5
(A)
77(02-3-22)
;1
0;
a
a
3
2
3 log
2
log
3
a
a
a
a
(A)
78(02-3-33)
log 128
log 64
log 32
16
log
log 8
log 4
log 2
4
log
1
4
log
1
4
log
1
4
log
1
4
log
1
4
log
1
4
log
1
4
4
4
4
2
4
4
128
64
32
16
8
4
2
28
2
2
7
6
5
4
3
2
4
2
log
2 )
2
2
2
2
log (2 2
=
14
2
28
(A)
79(02-4-38)
1
log 6
log 3
2
log
6
1
6
1
6
1
(B)
80(02-5-23)
;
3 log
1
2 log
; 1
log
log
2
2
8
4
b
a
b
a
5,1
2
3
log
log
log
2;
3
log
log
2
2
2
2
a
b
b
a
b
a
(A)
81(02-5-24)
1
27
1
3
1
3
3
3
3
log 3
3 3 3
log
27
27
1
1
(A)
82(02-6-20)
2;
1
3
1
1
3
1
27
1
9
1
3
1
4 ;
1
2
1
log16
4
1
128
4
1
128
4
1
log
,0 25
log
14
1
2
log
2
1
27
(C)
83(02-6-36)
3
6
3
7
8
3
6
2
6
7
3
2
log 7
6
log
log 2
3
log
216 log 343
log
256
1
729 log
log
11
3
51
3
16
3 log 6 log 7
3
6 log 3 log 2
8
6
7
3
2
(D)
84(02-8-12)
4;
7
log5
log5 7
b
b
2
4 2
1
2
1
log5 7
7
log5
b
b
(A)
85(02-8-13)
;
; lg3
lg2
b
a
b
a
2
1
lg3
2
lg2
1
9
lg
lg20
20
log9
(A)
86(02-8-15)
3
3
5
3
3
5
23
log 5 log 3
2
log
2
8
3
5
3log 5 log 3
2
3
3
3
5
3
log2 5
(A)
87(02-9-34)
49
7
3
9
2
log3 7 2
log3 7
;
5,0
6 3
1
6 50
1
49
6 1
1
log50 3
log50 3
(B)
88(02-9-58)
a
b
a
b
a
b
a
b
a
b
a
log
log
log
2
2
1
log
1
1
1
log
log
1
1
log
1
1
b
b
a
b
a
a
a
a
a
;
3;
; log
log
1
2 ; 2
1
log
1
1
b
b
b
a
a
a
8,0
3
2
3
1
log
2
log
1
)
(
log
)
log (
log
2
2
b
b
a b
ab
ab
a
a
a
a
a b
(E)
89(02-10-27)
;
; log 7
lg2
2
b
a
10
log
log 7
2
2
lg
10
log
log 28
lg2
lg28
lg2
56
lg
2
2
2
2
)
(3
3
2
1
2
b
a
ab
a
ab
a
a
a
b
a
(A)
90(02-10-73)
3
6 3
3
6
6
6
3
6
3
6
6
6
27
1
log
4
1
log
log 4
27
log
3
log 1
,0 25
log
2log 2
27
log
27
3 log 108
1
108
log
108
1
log
108
log
3
6
6
3
3
6
6
(A)
91(02-11-31)
2
5
log
10
5
log
2
1
2
2
5
log
2
5
5
log
2
2
5;
log
2
5
2
5
5
log
2
2
25
5
2
16
4
5 4
log2
5
log2
(C)
92(02-11-32)
;
2
6
log
1
3
log
2
2
a
x
2
6
log
1
3
log
2
2
bo’lsin.
a x
2
6
log
1
3
log
2
2
2
6
log
1
3
log
2
2
=
2
6
2
6
log
1
3
1
3
log
2
2
=
2;
1 1
log 2
2
log
2
2
a
x
x
a
2
2;
(E)
93(02-12-48)
13
,0
1300
lg
lg )5
(lg2
3
1300 lg ,013
lg
3lg5
3lg2
,0 75
4
3
10000
lg
3lg10
(E)
94(03-1-20)
;
5,0
5,0 ; log 3
2
log
11
5
;1
log 11
3
log
5
5
x
shuning uchun 3 x eng
katta son. (E)
95(03-2-20)
log 2
6
log
log 12
2
log
log 2
1
2log 2
1
2
6
2
3
3
2
6
2
3
3
=
6
log
log 6
12
log
6
log
log 2 log 6
12
log
2
3
2
6 2
log
3
3
2
3
2
3
6
3
6
log
log 2
log 2 1
6
log
6
log
log 2
log 2
6
log
2
3
3
3
3
2
3
2
3
3
3
12
6
log
log 2)
6 1(
log
6
log
log 2 log 6
6
log
2
3
3
3
2
3
3
3
3
1
6
log
6
log
2
3
2
3
(C)
96(03-2-25)
2
3
log
1
log
1
lg
1
p
4
log 100
log 2
log 5
log 10
(E)
97(03-3-33)
log 32
log 5
5
log
8
5 32
log
8
log25 32
2
8
=
3
5
2
log
5
23
(C)
98(03-4-32)
log8 ,0 36
log3 ,0 64
8
ln 3
y
0
ln1
,0 36)
ln( ,0 64
(C)
99(03-4-33)
log 32 18
3log 4
2log 8
2
8
4
log 32 18
log 64
64
log
2
8
4
24
5 18
2
3
(B)
100(03-5-39)
5 5 5 5
5
5
5
log log
y
4
log 5
log 5
log
4
5
4
5
5
5
(A)
101(03-6-59)
;
7,0
lg5
7
13
7
10
7,0
1
lg5
1
log5 10
(D)
102(03-9-19)
2
5
1
log
2 10
7
3
log
3
3
1
2
3
3
2
5
log
3
2 10
7
log
2 10
7
3
2 10
7
log
2
5
3
2 10
7
log
3
2
3
2
1
3
log
3
1
log
2
1
3
3
=-0,5 (D)
103(03-11-82)
log2 3
log3 2
2
;
3
y
x
bo’lsin. Har
ikki tenglikni 2 asosga ko’ra logarifmlaymiz.
log 2 log 3
log 3
log
2
3
3 2
log
2
2 x
log 3
2 log 3 log 3
log
2
2
2
3
;
log 3
log 2
log
2
2 3
log
2
2
y
bundan
0;
;
;
log
log
2
2
y
x
y
y x
x
0
2
3
log2 3
log3 2
;
1
1
2
3
log2 3
log3 2
(D)
1.14.3. Logarifmik tenglamalar.
1(96-6-55)
;
0
16
3
16;
3
2
3
log
log3
2
x
x
x
4
;
0
16
2
x
x
x
(C)
2(97-2-55)
24
19;
5
19;
4
5)
log4(
x
x
x
24-20=4 (C)
3(97-8-40)
;
0
25
4
25;
4
2
4
log
log4
2
x
x
x
5
;
0
25
2
x
x
x
(A)
4(98-2-35)
;
0
1
4
0
1
8
1
8
1
4
;1
8
2
3
3
1)
2 ( 3 4
log
x
x
x
x
x
x
x
x
x
0
1
4
0
1
8
2
2;
;
0
;
0
1
4
0
1
8
0
4)
(
3
3
2
1
3
2
x
x
x
x
x
x
x
x
x
x
x
x 2
chet ildiz.
Javob 0 va 2 (C)
5(98-6-30)
.?
;1
6;
3
2
2
1
2
x
x
x
x
Tenglamaning har ikki tomonini 2 asosga ko’ra
logarigmlaymiz.
log 6
log 3
log 6; log 2
3
2
log
2
2
2
2
2
2
2
x
x
x
x
;
log 6;
0;
log 6
3
log
2
2
1
2
2
2
x
x
x
x
log2 6;
2
x
(A)
6(98-9-34)
3 ;)
lg(
3)
2
lg(
2
x
x
x
13
;
3
0
3
2
0
3
;
0
3
0
3
2
3
3
2
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x
;
3
0
3
2
3
;
0
2
2
1
x
x
x
x
x
har ikki ildiz chet ildiz.
x
(D)
7(99-2-32)
2 ;
; 2
4
2
1
2
4
log
log4
x
x
16
1
2;
2; log
log
4
4
x
x
x
;
256
16
1
16
16
x
(E)
8(99-6-26)
0;
1
log log
log
2
2
18
x
2;
1
;1 log
1
log
log
2
2
2
x
x
4 ;
1
4;
1
x
x
(D)
9(99-6-28)
;
3log
)
log (54
2
3
2
x
x
;
log
)
log (54
3
2
3
2
x
x
0
54
27
;
0
0
54
54
3
3
3
3
3
x
x
x
x
x
x
x
bulardan
x 3
(D)
10(99-6-50)
0;
5
log
log
5
5
1
x
5
25;
5; 5
5
;1
5
log5
x
x
x
x
(E)
11(99-8-31)
;
3
3; 3
1
3
1
4
2 log 5
1
4
log 5
x
x
1
2
2 log
1
5
x
;
;1
2 log
1
5
x
2;
log 5
x
x 5
(D)
12(00-1-37)
0;
log log
log
2
4
8
x
4;
;1 log
log
log
2
2
2
x
x
x 16;
(C)
13(00-2-24)
7;
4log
2log 3
log
25
5
5
x
log 49;
log 9
log
5
5
5
x
log (9 49 ;)
log
5
5
x
441
9 49
x
(A)
14(00-3-28)
8 ;
lg
4
lg
8
27
9
4
1
x
x
log 4;
3
2
3
2
8
1)
3(
2
x
x
3;
2
3
2
3
3
2
x
x
2
;1
3
x
x
(C)
15(00-3-36) )
0;
log log
log
3
4
3
2
x
3;
;1 log
log
log
3
4
3
4
3
x
x
16
4;
43 ;
3
x
x
x
(B)
16(00-3-38)
lg ;
2
lg 1
2
lg 1
x
x
0
0
2
1
x
x
bundan
x 0
;
2 ;
1
2
2
2 ; 1
lg 1
2
2
lg 1
x
x
x
x
2
1
0;
1
0;
1
2
;1
2
2
1
2
2
x
x
x
x
x
x
Javob: 2
1 (B)
17(00-3-39)
100;
lg 1
x x
x>0; tenglamaning har
ikki tomonini 10 asosga ko’ra logarifmlaymiz.
t
x
x
x
x
x
2; lg
)1
lg100; lg (lg
lg
lg 1
bo’lsin.
;1
2;
0;
2
2;
)1
(
2
1
2
t
t
t
t
t t
10;
1
;1
100; lg
2;
lg
1
x
x
x
x
10
10
1
100
2
1
x x
(A)
18(00-3-41)
x
x
x
x
2
3
3
8
; 3
2
8
log 3
;
0)
(
3 ; 3
9
8
3
t
t
x
x
x
bo’lsin.
9 ;
8
t
t
0;
1
9;
0;
9
8
2
1
2
t
t
t
t
2
9;
3
x
x
(C)
14
19(00-8-6)
;
1
2
10
2
log3
2
3
log
x
x
x
x
2 ;
10
2
log 3
2
log 3
x
x
x
x
bundan
x1 1
;
2;
10
log
log
2
3
2
3
x
x
t
x
x
x
3
3
2
3
log
;0
8
2log
log
bo’lsin.
4;
2;
0;
8
2
2
1
2
t
t
t
t
81;
1
4;
9; log
2;
log
3
3
2
3
x
x
x
x
(A)
20(00-8-15)
1;
2log 3
7
9
log
1
2
1
2
x
x
1 ;
log 3
7
9
log
2
1
2
1
2
x
x
1 ;
3
7
9
2
1
1
x
x
6;
2 3
;1
2 3
9
7
9
1
1
1
1
x
x
x
x
3;
3
x1
2
;1
1
x
x
(A)
21(00-8-40)
8
3
3
3
2
3
3
log
log
log
log
x
x
x
x
;
log
log
36
3
8
3
2
3
x
x
x
x x
0;
27
0;
27
;
27
3
6
30
30
36
30
36
log3
x
x
x
x
x
x
3
27
27;
0;
6
6
x
x
x
(A)
22(00-10-82)
25;
2 )
(
45)2
log2 (
x x
x
5,0
0
1 ;
2
0
2
x
x
x
x
5;
5,4
25;
5,4 )
(
2
x
x
0
5,9
;
5,0
2
1
x
x
. Tenglama yechimga ega
emas. (A)
23(96-10-38)
t
x
x
x
2
2
2
2
log
;0
6
5log
log
bo’lsin.
2;
3;
0;
6
5
2
1
2
t
t
t
t
4;
2;
8; log
3;
log
2
2
1
2
x
x
x
x
32
4 8
2
1
x x
(C)
24(96-1-35)
t
x
x
x
lg
;0
2
lg
lg2
bo’lsin.
;1
2;
0;
2
2
1
2
t
t
t
t
10;
1
;1
100; lg
2;
lg
2
1
x
x
x
x
10
10
1
100
2
1
x x
(C)
25(96-9-86)
t
x
x
x
3
3
2
3
log
;0
3
4log
log
bo’lsin.
;1
3;
0;
3
4
2
1
2
t
t
t
t
3;
;1
27; log
3;
log
2
3
1
3
x
x
x
x
81
27 3
2
1
x x
(B)
26(98-3-33)
t
x
x
x
3
3
2
3
log
;0
2
3log
log
bo’lsin.
;1
2;
0;
2
3
2
1
2
t
t
t
t
3;
;1
9; log
2;
log
2
3
1
3
x
x
x
x
12
3
9
2
1
x x
(C)
27(98-6-24)
t
x
x
x
2
2
2
2
log
;0
1
4log
log
bo’lsin.
5;
2
5;
2
0;
1
4
2
1
2
t
t
t
t
;
2
5;
2
log
5
2
1
2
x
x
;
2
5;
2
log
5
2
2
2
x
x
x1 x2
16
2
2
2
4
5
2
2 5
(C)
28(98-10-80)
t
x
x
x
2
2
2
2
log
;0
3
4log
log
bo’lsin.
;1
3;
0;
3
4
2
1
2
t
t
t
t
2;
;1
8; log
3;
log
2
2
1
2
x
x
x
x
10
8
2
2
1
x x
(B)
29(98-12-105)
;)
(3 13
)3
13| log (
|
2
x
x
x
;
0
13
0
13)
3(
3)
13)log (
(
)1
2
x
x
x
x
13
;
13
0
3)
3)
13)(log (
(
1
2
x
x
x
x
13
8
31
;
13
3
3)
(
log
2
2
x
x
x
15
;
0
13
0
13)
3(
)3
13)log (
(
)
2
2
x
x
x
x
11
;
13
3
)3
(
log
2
2
x
x
x
24
13
11
2
1
x x
(D)
30(99-3-20)
lg30;
1
3
lg 2
5
lg
x
x
5
;
5,1
5
0;
3
2
0
5
x
x
x
x
x
;
;1
lg30
3
5 2
lg
x
x
lg3;
)3
5)(2
(
lg
x
x
9
3)
5)(2
3 ; (
)3
5)(2
(
2
2
x
x
x
x
;
5
2
1
6;
0;
6
13
2
2
1
2
x
x
x
x
;
Javob: x=6 (B)
31(99-6-55)
4;
2
log
2
log
2
x
x
0;
;1
x
x
4;
2log
2log
2
2
x
x
4;
4log
2
x
;1
log2
x
x=2 (A)
32(99-8-32)
;1
)lg
lg
(lg10
;1
)lg
10
lg(
2
2
x
x
x
x
;1
2lg )lg
(lg10
x
x
t
x
x
x
lg
;0
1
lg
2lg
2
;1
2 ;
1
0;
1
2
2
1
2
t
t
t
t
10.
1
;1
10; lg
2 ;
1
lg
2
1
x
x
x
x
Kichik ildiz 0,1 (B)
33(00-1-47)
;0
3
2log
log
2
2
2
2
x
x
.
t
x
x
x
2
2
2
2
log
;0
3
4log
log
bo’lsin.
;1
3;
0;
3
4
2
1
2
t
t
t
t
2;
;1
8; log
3;
log
2
2
1
2
x
x
x
x
10
2
8
2
1
x x
(D)
34(00-4-40)
lg (100 );
6
lg (10 )
lg
2
2
2
x
x
x
lg ) ;
(lg100
6
lg(10 ))
lg(10 ))(lg
(lg
x 2
x
x
x
x
lg ) ;
(2
6
lg )
lg10
(lg
10
lg
x 2
x
x
x
x
);
lg
4lg
(4
6
)1
(2lg
2 x
x
x
;
lg
4lg
4
6
1
2lg
2 x
x
x
t
x
x
x
lg
;0
3
2lg
lg2
bo’lsin.
3;
;1
0;
3
2
2
1
2
t
t
t
t
;
1000
1
3;
10; lg
;1
lg
2
1
x
x
x
x
,0 01
1000
1
10
2
1
x x
(D)
35(97-12-54)
;1
3)
log (
2)
log (
2
2
x
x
2;
3;
2
0;
3
0
2
x
x
x
x
x
2;
)3
2)(
;1 (
)3
2)(
log2 (
x
x
x
x
0;
4
;1
0;
4
5
2
1
2
x
x
x
x
9
)1
(
8
8
x
(B)
36(98-11-45)
2;
log
2
2 log
log
4
2
x
x
x
(4 )
log
1
log (2 )
log
1
4
2
2
x
x
x
;
;)
log (4
log (2 )
log
2
2
2
x
x
x
;
log
2
)
log
1(
log
2
2
2
x
x
x
2;
; log
log
2
log
log
2
2
2
2
2
2
x
x
x
x
;
2
2;
log
2
1
2
x
x
;
2
2;
log
2
2
2
x
x
;1
2
2
2
2
2
2
x
x
(A)
37(99-3-21)
;1
12) log 2
log4 (
x
x
0
;1
x
x
;1
log
12)
2 log (
1
2
2
x
x
;1
12)
2 log (
1
x x
;
12
2;
12)
log (
x2
x
x x
0;
3
4;
0;
12
2
1
2
x
x
x
x
(A)
38(00-8-39)
36;
log
log
log
log
16 5
6 5
4 5
5
x
x
x
16
36;
16log
6log
4log
2log
5
5
5
5
x
x
x
5
2 ;
1
36; log
72log
5
5
x
x
x
(A)
39(00-10-40) )
3;
log
3
3 log
log
9
3
x
x
x
(9 )
log
1
log (3 )
log
1
3
3
3
x
x
x
;
;)
log (9
log (3 )
log
3
3
3
x
x
x
;
log
2
)
log
1(
log
3
3
3
x
x
x
2;
; log
log
2
log
log
2
3
3
2
3
3
x
x
x
x
;
3
2 ;
log
2
1
3
x
x
;
3
2 ;
log
2
2
3
x
x
;1
3
3
2
2
2
2
x
x
(C)
40(97-1-59)
10;
25lg
lg25
x
x
;
25lg
lg25
x
x
;1
5; 2lg
5; 5
10; 25
25
2
2lg
lg
lg
x
x
x
x
10;
2 ;
1
lg
x
x
(C)
41(97-6-59)
6;
9lg
lg9
x
x
;
9lg
lg9
x
x
;1
3; 2lg
3; 3
6; 9
9
2
2lg
lg
lg
x
x
x
x
10;
2 ;
1
lg
x
x
(C)
42(97-7-35)
;1
)1
lg(
x
x
tenglama ildizlarining
sonini funksiya grafiklarini chizish orqali topamiz.
;
1
)1
lg(
x
y
x
y
Funksiya grafiklari ikki nuqtada kesishgan.
Tenglama ikki ildizga ega. (B)
43(97-10-35)
3;
)1
ln(
x
x
tenglama ildilaining
sonini funksiya grafiklarini chizish orqali topamiz.
;
3
)1
ln(
x
y
x
y
Funksiya grafiklari ikki nuqtada kesishgan.
Tenglama ikki ildizga ega. (B)
44(00-2-22)
3
972
2
2; 3
)
(
log
972
2
3
3
y
x
y
x
y
x
y
x
;
972;
2
972; 27 3
2
3
;
3
972
2
3
3
y
y
y
y
y
x
y
x
10
5 2
5;
2;
36;
6
xy
x
y
y
(C)
45(00-4-43)
0
)
8
1
5)( 2
3
ln(
7)
12 lg(2
7
7
2
7
5
2
x
x
x
x
x
x
x
x
;
1)
3 ;
2
;
0
8
1
2
0
5)
3
ln(
x
x
x
x
x
2)
;
0
4)
3)(
(
8
5,3
5,0
3
5
;
0
12
0
8
0
7
2
0
1
2
0
5
3
2
x
x
x
x
x
x
x
x
x
x
x
x
5,3 ;8
x
a)
0;
6
5
;1
7
5
7;
7
2
2
7
2 5
x
x
x
x
x
x
5,3 ;
2
5,3 ;
3
2
1
x
x
17
b)
5,3
4
5,3 ;
3
0;
12
2
1
2
x
x
x
x
d)
4
;1
7
0; 2
7)
lg(2
x
x
x
;
Tenglama bitta ildizga ega. (B)
46(01-1-22)
2 ;
1
39)
3
lg(
5)
2
lg(
2
x
x
;
0
39)
3
lg(
0
39
3
0
5
2
2
2
x
x
x
39 ;)
lg(3
5)
2lg(2
2
x
x
39 ;)
lg(3
5)
lg(2
2
2
x
x
39;
3
25
20
39; 4
3
5)
(2
2
2
2
2
x
x
x
x
x
4;
16;
0;
64
20
2
1
2
x
x
x
x
(D)
47(01-1-23)
;1
3
log
1
log
x
x x
3
;1
log 3
;1
3
log
1
;1
3
log
1
2
x
x
x
x
(B)
48(01-2-66)
3 ;
2cos5
3)
(2
log4
x
2;
3
4;
3
;1 2
3)
log4 (2
x
x
x
1
4;
3
x
x
(A)
49(01-2-73)
;
10
5 lg
3
5
lg
x
x
x
tenglamaning har ikki
tomonini 10 asosga ko’ra logarifmlaymiz.
lg ;
5
lg
3
5
lg
;
lg10
lg
5 lg
3
5
lg
x
x
x
x
x
x
3lg ;
15
5lg
lg2
x
x
x
t
x
x
x
lg
;0
15
2lg
lg2
bo’lsin.
5;
3;
0;
15
2
2
1
2
t
t
t
t
;
10
1
5;
10 ; lg
3;
lg
5
2
3
1
x
x
x
x
,0 01
100
1
10
1
10
5
3
2
1
x x
(E)
50(01-3-26)
10;
1
1 ; 3
1
lg 3
3
4
2
3
4
2
x
x
x
x
x
x
4;
3
4
2;
3
4
3 ;
3
2
2
2
3
4
2
x
x
x
x
x
x
x
x
x
;
3
0
3
4
2
x
x
x
x
6;
2;
0;
12
8
12;
4
4
2
1
2
2
x
x
x
x
x
x
x
8;
6
2
2
1
x x
(C)
51(01-5-10)
;1
lglg2
10
lg 4
lg
x
x
4;
0
0 ;
4
;
0
0
4
x
x
x
x
x
;
10
2
lg
10
4
;
10
lg lg2
10
4
lg
2
x
x
x
x
0;
lg2
4
lg2;
4
2
2
x
x
x
x
;
2
4lg2
16
4
2
,1
x
Bu ildizlar (0;1) oraliqqa
tegishli bo’lgani uchun. Berilgan tenglama ikki
ildizga ega. (A)
52(01-5-11)
4 ;
3
log
log
log
4
2
x
x
x
a
a
a
4
3
4 log
4 ; 3
3
4 log
1
2 log
1
log
x
x
x
x
a
a
a
a
a
x
a x
;1
log
(A)
53(01-5-12)
0;
;1
3;
log ( 2 1)
x
x
x
x x
0;
2
2;
4;
3;
1
2
1
2
2
x
x
x
x
(A)
54(01-7-25)
0;
))
2lg(1
lg(3
x
;1
)1
2; lg(
)1
;1 2lg(
)
2lg(1
3
x
x
x
9,0
10;
1
1
x
x
(D)
55(01-7-26)
2;
|1
;1 |
|1
log2 |
x
x
;1
;2
1
3;
2;
1
2
1
x
x
x
x
(E)
56(01-8-33)
;
2,0
lg
lg ,0 2
x
x
Istalgan musbat son
tengsizlikning yechimi bo’ladi, shuning uchun 5 ga
karrali eng kichik ildiz x=5.
18
11
6
6
6
log611
6
lg
11
lg
lg6
6 5) 1
lg(
(C)
57(01-9-1)
4;
) log
(9
log
2
3
2
x
x
x
4;
) log
log
9
(log
2
3
2
x
x x
x
2;
log
1
log
1
2;
)1 log
3
(log
2
3
3
2
3
x
x
x
x
log3 x t
bo’lsin.
2;
2;
1
1
2
2
t
t
t
t
2;
;1
0;
2
1
1
2
t
t
t
t
3;
;1
log
1
3
x
x
9
1
2;
log
2
3
x
x
9 ;
31
9
1
3
2
1
x x
(A)
58(01-9-9)
9;
3 log3
x
x
tenglamaning har ikki
tomonini 3 asosga ko’ra logarifmlaymiz.
2;
)log
log
log 9; (3
log
3
3
3
log3
3
3
x
x
x
x
;
log3
x t
bo’lsin.
0;
2
3
2;
2; 3
)
(3
2
2
t
t
t
t
t t
2;
;1
2
1
t
t
9;
2;
3; log
;1
log
2
3
1
3
x
x
x
x
3 3
9 3
(A)
59(01-9-22)
1
;0
1
0;
)1
3lg(
)
169
lg(
3
x
x
x
x
0;
)1
(
0; lg169
)1
lg(
)
169
lg(
3
3
3
3
x
x
x
x
;1
3
3
169
;1
)1
(
169
2
3
3
3
3
x
x
x
x
x
x
0;
56
0;
168
3
3
2
2
x
x
x
x
7;
;1
8
2
1
x
x
(A)
60(01-9-36)
;)1
log (2
;)1
log (
;)1
log (
3
3
3
x
x
x
1
0;
1
x
x
.
Bu sonlar arifmetik progressiyaning ketma-ket
hadlari bo’lsa quyidagi tenglik bajariladi.
;)1
log (2
)1
log (
)1
2log (
3
3
3
x
x
x
)1 ;
1)(2
log (
)1
log (
3
2
3
x
x
x
;1
2
2
1
2
;)1
1)(2
(
)1
(
2
2
2
x
x
x
x
x
x
x
x
5
;1
0
0;
5)
(
0;
5
2
1
2
x
x
x x
x
x
(E)
61(01-9-41)
5 )
lg(2
2)
lg(5
x
x
;
x
x
x
x
x
;
5
2
5
2
0;
5
2
0
2
5
(B)
62(01-9-44)
0;
13)
8
log (
13)
5
(
log
2
2
7
1
2
2
7
x
x
x
x
Bu tenglik bajarilishi uchun har bir qo’shiluvchi 0
ga teng bo’lishi kerak.
;
1
13
8
1
13
5
0 ;
13)
8
(
log
0
13)
5
(
log
2
2
2
2
7
1
2
2
7
x
x
x
x
x
x
x
x
2 ;
;6
7
;
2
2
1
2
1
x
x
x
x
x=2 (B)
63(01-10-29)
;
10
2
2
lg
x
x
x
tenglamaning har ikki
tomonini 10 asosga ko’ra logarifmlaymiz.
;2
lg
2
2
lg
;
10
2lg
lg
2
lg
;
10
lg
lg
2
2
2
lg
x
x
x
x
x
x
x
x
0;
2
0; lg
2)
0; (lg
4
4lg
lg
2
2
x
x
x
x
100;
2;
lg
x
x
(B)
64(01-10-32)
515;
log
2
3
3
2
x
x
x=3 berilgan
tenglamaning ildizi. Endi tenglamaning boshqa
ildizi yo’qligini ko’rsatamiz.
3
3
2
log
515
;
2
x
y
y
x
; birinchi funksiya o’suvchi
ikkinchi funksiya esa kamayuvchi. Shu sababdan
tenglama x=3, yagona ildizga ega. (B)
65(01-11-29)
0;
2
0; 3
;1
4;
2)
log (3
2
2
x
x
x
x
x
;
0;
2
3
;
2
3
2
2
4
4
2
t
x
x
x
x
x
19
;1
2;
0;
2
3
2
1
2
t
t
t
t
2
0;
2
2;
;1
;1
2
2
x
x
x
x
x
;
tenglama yagona ildizga ega. (D)
66(01-11-33)
;
26 3
3
log
2
3
x
x
x
26;
0; 3
26 3
32
x
x
x
0;
27)
0; 3 (3
27 3
3 ; 3
26 3
3
2
2
x
x
x
x
x
x
x
3
27;
0; 3
3
x
x
x
(D)
67(02-10-60)
5,0
log 2
2,0
;
5,0
32
1
log
2,0
5
x
x
;
4;
2;
2 ;
1
5,0 ; log 2
2
log
2
1
x
x
x
x
(D)
68(02-2-24)
0;
6 ;
log
2
log
3
3
3
x
x
x
x
x
0;
2
6
log
log
3
3
3
x
x
x
x
0;
2)
1)(log
0; (3
)1
2(3
)1
(3
log
3
3
x
x
x
x
x
3;
1
0;
1
)1 3
x
x
9
2;
2) log
3
x
x
(A)
69(02-3-35)
0 ;
1
4 3
;1
2
)1
log3 (4 3
x
x
x
;
3
0;
1
4 3
3 3
;
3
1
3
4
2
2 1
t
x
x
x
x
x
3;
1
;1
0;
1
4
3
2
1
2
t
t
t
t
;1
;
3
0; 3
;1
3
1
x
x
x
x
1
0
1
2
1
x x
(A)
70(02-3-36)
0;
4 ;
1
;1
;1
4
log
2
log
4
x
x
x
x
x
;
log
;1
log
2
2
log
1
;1
4
log
1
log
1
2
2
2
4
2
t
x
x
x
x
x
0;
2
;
2
2
2
;1
2
2
1
2
2
t
t
t
t
t
t
t
t
;
2
1
;1
4; log
2;
;1 log
2;
2
2
1
2
2
1
x
x
x
x
t
t
2
2
1
4
2
1
x x
(A)
71(02-3-31)
6 ;
lg 3
2 lg3
1
1
x
x
0
6; 3
3
3
6 ;
lg 3
lg3
2
1
1
2
1
1
2
1
t
x
x
x
x
x
bo’lsin.
;0
2
3;
0;
6
6;
2
1
2
2
t
t
t
t
t
t
2
1
;1
2
1
3;
3 2
1
x
x
x
(A)
72(02-5-25)
lg25;
2
7)
lg(2
5,0
11)
lg(
x
x
25 ;
lg100
7
2
lg
11)
lg(
x
x
7 ;
4 2
1
;
4
7
2
11
4;
lg
7
2
11
lg
x
x
x
x
x
x
7 ;)
11 16(2
22
;
7
4 2
11)
(
2
2
2
x
x
x
x
x
0;
9
10
112;
32
121
22
2
2
x
x
x
x
x
10
2
1
x x
(D)
73(02-5-27)
3
)
(
log
2
1
2 log
1
;
3
)
(
log
2
1
log
3
2
3
3
2
9
xy
y
x
xy
y
x
;
;
6
2log
log
2
1
2 log
1
2log
;
3
log
log
1
log
log
3
3
3
3
3
3
3
2
3
y
x
y
x
y
x
y
x
tenglamalarni ayiramiz.
3;
9;
2;
5; log
2 log
5
3
3
x
y
y
y
12
3
9
x y
(C)
74(02-6-34)
;
10
2
2lg
x
x
x
tenglamaning ha ikki
tomonini 10 asosga ko’ra logarifmlaymiz.
2lg ;
1
lg
; 2lg
lg10
lg
2
2lg
x
x
x
x
x
x
t
x
x
x
0; lg
1
2lg
2lg
2
bo’lsin.
;
2
2
1
;
2
2
1
0;
1
2
2
2
1
2
t
t
t
t
;
10
;
2
2
1
; lg
10
;
2
2
1
lg
2
2
1
2
2
2
1
1
x
x
x
x
10
10
10
2
2
1
2
2
1
2
1
x
x
(B)
20
75(02-6-35)
84;
log
3
3
2
2
x
x
x=2 berilgan
tenglamaning ildizi. Endi tenglamaning boshqa
ildizi yo’qligini ko’rsatamiz.
3
2
2
log
84
;
3
x
y
y
x
; birinchi funksiya o’suvchi
ikkinchi funksiya esa kamayuvchi. Shu sababdan
tenglama x=3, yagona ildizga ega. (B)
76(02-7-6)
45;
3
3
2
log 73
log 7
x
x
;
3
3
log7
log7
x
x
9;
45; 3
5 3
log 7
log 7
x
x
49;
2;
log7
x
x
(A)
77(02-8-16)
3;
)
1(
log
)
log (3
5,0
2
x
x
3
1;
3
0 ;
1
0
3
x
x
x
x
x
;
;3
)
)(1
3(
;3 log
)
log 1(
)
3(
log
2
2
2
x
x
x
x
1
;3
5
0;
5
4
;8
3
3
2
1
2
2
x
x
x
x
x
x
x
;
6
)1
(
5
ni qo’shsak. (A)
78(02-10-69)
2log 12;
16 )
log (2
4
2
2
x
x
0
12; 4
16
log 12; 4
16 )
log (2
2
2
2
t
x
x
x
x
x
;
3;
0;
4
0;
12
12;
2
1
2
2
t
t
t
t
t
t
log 3
3;
4
4
x
x
(A)
79(02-10-71)
;
72
3
2
2
;
72
3
2
1
)
(
log
1
1
2
y
x
y
x
y
x
y
x
72;
3 3
72; 4 2
3
2
;
72
3
2
2
1
2
1
y
y
y
y
y
x
y
x
3
3 1
3;
;1
6;
6
xy
x
y
y
(A)
80(02-11-33)
3;
log (4 )
2
log
2
2
2
x
x
3;
)
log
(lg 4
log 2
log
2
2
2
2
2
x
x
3;
)
log
(2
1
log
2
2
2
x
x
3;
log
2
1
2log
log
2
2
2
2
x
x
x
t
x
x
x
2
2
2
2
0; log
4
3log
log
bo’lsin.
;1
4;
0;
4
3
2
1
2
t
t
t
t
2 ;
1
;1
16; log
4;
log
1
2
1
2
x
x
x
x
8
2
1
16
2
1
x x
(D)
81(02-11-34)
6;
log
log
log
3
1
3
x
x
x
x
6;
log
2log
log
3
3
x
x
x
x
4;
2log
3
x
9
2;
log3
x
x
;
18
4
9
9
4
2
2
x x
(D)
82(02-12-49)
;
26 3
3
log
2
3
x
x
x
26;
0; 3
26 3
32
x
x
x
0;
27)
0; 3 (3
27 3
3 ; 3
26 3
3
2
2
x
x
x
x
x
x
x
3
27;
0; 3
3
x
x
x
(D)
83(02-12-50)
0;
;
lg96
)
lg(
2
lg
2
)
lg(
2
2
x
xy
y
x
;
96
2
100
;
lg96
)
2
lg(
100
2
2
2
2
xy
y
x
xy
y
x
14
196;
)
(
2
y
x
y
x
(B)
84(03-1-11)
5 ;
4
;1
0;
2;
4)
log (5
x
x
x
x
x
;1
4;
0;
4
5
;
4
5
2
1
2
2
x
x
x
x
x
x
(B)
85(03-9-20)
2;
19)
19
(5
log
2
5 2
x
x
x
;
0
2
5,2
;
0
19
19
5
1
2
5
0
2
5
2
D
x
x
x
x
x
x
2 ) ;
(5
19
19
5
2
2
x
x
x
0;
6
;
4
20
25
19
19
5
2
2
2
x
x
x
x
x
x
2;
3;
1
x
x
tenglama 1 ta ildizga ega, demak
m=1.
3
4
3
2
2 1
2
2
0
x
m
(C)
86(03-11-78)
0;
)log
(2
2
x
k
x
1
0;
log2
x
x
; k qanday qiymat qabul qilishidan
qat’iy nazar x=1 tenglamaning ildizi bo’ladi.
21
x=1 bo’lsa k=2 boladi. Tenglama boshqa ildizga
ega bo’lmasligi uchun
x2 0
bo’lishi kerak.
0;
0;
2
0;
2
2
k
k
x
k
x
Javob:
2
0;
k
k
(E)
87(03-5-28) Izlanayotgan sonlar
2
1
1
1
;
;
b b q b q
bo’lsin shartga asosan
1
0;
1
q
b
.
;
9
)
log (
)
log (
log
42
2
1
2
1
2
1
2
2
1
1
1
b q
b q
b
b q
b q
b
;
2
(
42
)
1(
;
9
)
(
log
42
9
3
3
1
2
1
3
3
1
2
2
1
1
1
q
b
q
q
b
q
b
b q
b q
b
8 ;
42
)
1(
;
8
42
)
1(
1
2
1
1
2
1
q
b
q
q
b
q
b
q
q
b
21 ;
4
4
4
4 ;
21
1
2
2
q
q
q
q
q
q
4
;1
4
1
0;
4
17
4
2
1
2
q
q
q
q
(A)
88(03-7-21)
;2
2)
(2
log
6)
(4
log
5
5
x
x
2
2
6
4
;
0
2
2
0
6
4
x
x
x
x
t
x
x
x
x
x
;5 2
2
2
6
4
;2
2
2
6
4
log 5
bo’lsin.
0;
4
5
10;
5
6
5;
2
6
2
2
2
t
t
t
t
t
t
;1
2; 2
4;
;1 2
4;
1
1
x
x
x
t
t
(C)
89(03-2-4)
;)
lg(10
lg
3;
lg
lg
2
2
4
3
2
2
4
b
a
a
b
b
a
a
b
10;
10 ;
;
10
;
10
3
3
3
4
2
4
3
2
a
a
a
a
b
a
a
b
100
;
10
2
4
b
b
;
10 100 110
a b
(D)
90(03-4-34)
;|
3log | 3
)
3
(
)
(2
log
4
3
2
4
x
x
x
0
2)
2; (
2
x
x
bo’lgani uchun
0
)
(3
x 3
bo’lishi kerak, demak 3-x>0, x<3 ; |3-x|=3-x;
;
)
log (3
)
3
(
)
(2
log
3
4
3
2
4
x
x
x
) ;
3
(
1
)
3
(
)
2
(
3
3
2
x
x
x
3
;1
;1
)
(2
2
1
2
x
x
x
chet ildiz.
26
1 27
27
x
(C)
91(03-8-47)
0;
4
lg5; 2
4)
lg(2
x
x
x
x
x
x
;
10
10
lg5
4)
lg(2
x x
x
x
;
10
10
4
2
lg5
x
x
x
x
;
5
10
4
2
x
x
x
x
4
2 ;
4
2
x
x
x
x
(A)
92(03-11-13)
;
2
7
3log2 7
2
9
2 5
2
x
x
2 ;
3
2
9
5
; 2
2
7
2
2
3
log2 7
2
9
2 5
2
x
x
x
x
5,1
;4
;0
12
5
;3 2
9
5
2
2
1
2
2
x
x
x
x
x
x
(E)
93(03-7-38)
0;
2
log 9
log
2
3
x
x
0 ;
log
;1 2
0;
3
x
x
x
0;
2
log
2 log
1
2
2
3
3
x
x
log3 x t
bo’lsin.
0;
2
2
0 ; 2 2
2
2
1
2
2
t
t
t
t
0)
2
(;
; 2
2
0;
2
2
t
t
t
t
t
t
t
;
;1
0;
2
0;
2
2
1
2
t
t
t
t
3
1
;1
log3
x
x
(A)
94(03-4-36)
lg1000;
lg
3 ;
log
1000
3
lg
lg
y
x
x
x
x
y
y
y
3;
3; 3lg
lg
3; lg
lg
lg
2
3
3
y
y
y
y
x
x
y
10;
1
10;
;1
3; lg
lg
2
1
2
y
y
y
y
(С)
95(03-12-76)
162;
3
log3
2
log3
x
x
x
x 0;
22
162;
162;
3
log3
log3
log3
3
log
log3
x
x
x
x
x
x
x
x
log 81;
81; log
161;
2
3
3
log
3
log3
log3
x
x
x
x
x
x
2;
4; log
4; log
log
log
3
2
3
3
3
x
x
x
x
1
9
9 1
9 ;
1
9;
2
1
2
1
x
x
x
x
(С)
96(03-3-34)
;1
5
log
25
log
2
2
,0
2
,0 2
x
x
;1
log 5
25
log
2
5
2
5
x
x
1
log 5)
(log
log 25)
(log
2
5
5
2
5
5
x
x
t
x
x
x
5
2
5
2
5
;1 log
)1
(log
2)
(log
bo’lsin.
;1
1
2
4
4
;1
)1
(
2)
(
2
2
2
2
t
t
t
t
t
t
2;
;1
0;
2
3
0;
4
6
2
2
1
2
2
t
t
t
t
t
t
25;
2;
5; log
;1
log
2
5
1
5
x
x
x
x
125
25
5
2
1
x x
(B)
97(03-9-21)
0;
;
4
1
;1
;1
4
log
1
4
log
2
4
x
x
x
x
x
x
;1
log
4
log
4
log
2
4
4
4
x
x
x
;1
log
log
1
log
1
2
4
4
4
x
x
x
log4 x t
bo’lsin.
0;
)
1(
1
1
;1
1
1
2
2
t
t
t
t
t
t
0;
1
1
1
)
0; 1(
)
)(1
1(
1
1
t
t
t
t
t
t
t
;1
0;
1
1
t
t
0;
1
)
1(
1
0;
1
1
1
0;
1
1
1
2
t
t
t
t
t
t
2;
0;
0;
2
0;
1
2
3
2
2
2
t
t
t
t
t
t
t
;
16
1
2;
log
;1
0;
4; log
;1
log
4
4
4
x
x
x
x
x
x
16
81
16
1
1
4
3
2
1
x
x
x
(С)
1.14.4. Logarifmik tengsizliklar.
1(96-3-87)
2 ;
4
log
log
2
3
2
x
x
y
;
1
2
4
0
2
4
;
0
2
4
log
0
2
4
2
2
2
3
2
x
x
x
x
x
x
x
x
0
3
4
;1
2
4
;
1
2
4
0
2
4
2
2
2
2
x
x
x
x
x
x
x
x
;
3;1
0;
)1
3)(
(
x
x
x
(B)
2(96-7-33)
0;
8
4
1
4
log
2
1
x
x
0;
1
8
4
1
4
;1
8
4
1
4
;
0
8
4
1
4
1
8
4
1
4
x
x
x
x
x
x
x
x
2;
0;
8
0; 4
8
4
9
0;
8
4
8
4
1
4
x
x
x
x
x
x
x ;2
(E)
3(97-1-24)
4;
5)
(
2log
5
log
3
3
1
x
x
4;
5)
4log (
5)
2log (
3
3
x
x
5 ;
14
0 ;
5
9
5
2;
5)
log3 (
x
x
x
x
x
;514
x
(B)
4(97-1-56)
;1
2 )
log5 (5
x
;
5,2
0
5;
2
0
2
0;
2
5
5
2
5
x
x
x
x
x
x
5,2
x ;0
(D)
5(97-3-33)
0;
5,1
3
3
log 3
x
x
0;
1
5,1
3
3
;1
5,1
3
3
;
0
5,1
3
3
1
5,1
3
3
x
x
x
x
x
x
x
x
5,0 ;
0;
5,1
0; 3
5,1
3
5,1
0;
5,1
3
5,1
3
3
x
x
x
x
x
x
5,0 ;
x
(A)
6(97-4-16)
lg ;
2
x
y
y 0;
23
100
2;
2; lg
lg
0;
lg
2
x
x
x
(A)
7(97-6-24)
3 ;
4
2 )
log (3
2 )
(3
log
8
1
2
x
x
3 ;
4
2 )
3 log (3
1
2 )
(3
log
2
2
x
x
1
2 )
log (3
3 ;
4
2 )
3 log (3
4
2
2
x
x
5,0
;1
2
2;
2
3
x
x
x
;
5,0;
x
(A)
8(7-7-33)
0;
1
5
5
log
5
2
x
x
0;
1
1
5
5
;1
1
5
5
;
0
1
5
5
1
1
5
5
x
x
x
x
x
x
x
x
;
2,0
0;
1
0; 5
1
5
1
0;
1
5
1
5
5
x
x
x
x
x
x
2,0 ;
x
(B)
10(97-11-24)
2;
3
2)
log (
2
log
9
3
1
x
x
2 ;
3
2)
2 log (
1
2)
log (
3
3
x
x
;1
2)
2 ; log (
3
2)
2 log (
3
3
3
x
x
;
2
1
0 ;
2
3
2
x
x
x
x
1;2
x
(D)
11(97-12-53)
0;
2
1
4)
(
5,0
log
x
x
;4
0;
0;
4)
(
x
x x
. Eng kichik
musbat yechim 5 (C)
12(98-2-37)
;)1
(6
log
)1
2
(
log
2
2
,0
2
4
,0 2
x
x
x
0;
4)
(
0;
4
;1
6
1
2
2
2
2
4
2
2
4
x
x
x
x
x
x
x
;2 0
0;2
0;
2)
2)(
2 (
x
x
x
x
. Manfiy
yechimlar.
;2 0
(B)
13(98-3-32)
0;
log 12
)
log (3
5
5
x
;
3
9
0 ;
3
12
3
log 12;
)
log (3
5
5
x
x
x
x
x
3
9
x
. 11 ta butun son tengsizlikni
qanoatlantiradi. (D)
14(98-4-30)
;1
5,1
5
2
9
4
log3
x
x
;
0
5
2
5,7
3
9
4
0
5
2
5,7
3
9
4
;
0
5,1
5
2
9
4
3
5,1
5
2
9
4
x
x
x
x
x
x
x
x
x
x
0 ;
5,2
(
7
5,1
0
5,2 )
16 5, )(
(
;
0
5
2
5,1
7
0
5
2
5,
16
x
x
x
x
x
x
x
x
14;16 5,
3
x
. 16 ta butun son tengsizlikni
qanoatlantiradi. (A)
15(98-9-35)
20 ;)
(
log
4)
(2
log
2
3,0
2
3,0
x
x
0;
4)
4)(
0; (
16
20;
4
2
2
2
2
x
x
x
x
x
4;4
x
.
0
2
4
4
(E).
16(98-10-79)
0;
log 7
)
log (4
2
2
x
;
4
3
0;
4
7
4
;7
log
)
(4
log
2
2
x
x
x
x
x
4
3
x
. 6 ta butun son tengsizlikni
qanoatlantiradi. (A)
17(98-2-33)
2;
28
27
9
log
2
4
3 2 5
x
x
x
;1
5
0; 3
5
3
2
2
x
x
5) ;
(3
28
27
9
2
2
2
4
x
x
x
25;
30
9
28
27
9
2
4
2
4
x
x
x
x
0;
)1
1)(
0; (
1
0;
3
3
2
2
x
x
x
x
1;1
x
. Butun yechim 0 (E)
18(99-3-17)
0;
log log
log
5
3
1
2
x
24
3
5
5
5
3
1
5
3
1
5
5
5
1
0
;
3
1
log
1
log
1
0
;
1
log
log
0
log
log
0
log
0
x
x
x
x
x
x
x
x
x
x
x
x
;
x ;1 3 5
(E)
19(99-6-9)
3;
)1
log2 (2
x
5
,4;5,0
;
5,0
5,4
0;
1
2
8
1
2
x
x
x
x
x
. Eng katta
butun yechim 4 (D)
20(00-3-40)
;)
log (5
)1
log (
4
4
x
x
;1 2
;
5
1
2
;
0
5
0
1
5
1
x
x
x
x
x
x
x
x
(E)
21(00-4-41)
;1
2)
log 2 (
x
x
1)
;
2
0
)1
2)(
(
0
)1
1)(
(
;
2
0
2
0
1
;
0
2
2
1
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x
x
x
2;
2; 1
x
2)
;
0
2
0
)1
2)(
(
0
)1
1)(
(
;
0
2
0
2
0
1
;
0
0
2
2
1
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
0 1;
;1 0
x
. Har ikkovidan
;2
1;0
0;1
;2 1
x
(C)
22(00-4-42)
,2 4375
.2 25
2
,2 25
3
2
,2 25; 3
2
2
x
x
x
;
,0 8125
2
.2 25
,2 25
2
2
2
x x
,
)2
log (
2 )
log (3
2
2
x
x
x
x
c
c
tengsizlik
x=2,25 da to’g’ri bo’lishi uchun 0<c<1 bo’lishi
kerak, chunki 2,4375>0,871.
;
2
2
3
0
2
0
2
3
2
2
2
2
x
x
x
x
x
x
x
x
;
0
5
3
2
0
2
0
3
2
2
2
2
x
x
x
x
x
x
5,2
;2
;
0
)1
5,2 )(
(
0
)1
2)(
(
0
)1
3)(
(
x
x
x
x
x
x
x
(C)
23(00-7-34)
6;
)3
(
2
2
3)
log2(
x
x
;
3
0
)5
(
;
3
0
5
;
0
3
6
9
6
3
2
2
x
x
x
x
x
x
x
x
x
x
5;3
x
. Tengsizlikning eng kichik yechimi 4 va u
15 da 11 ta kam. (C)
24(00-9-22)
13) ;
log (
17)
(
log
8
5
1
8
5
1
x
x
13 ;|
17 | 8log |
log |
8
5
1
5
1
x
x
13 ;|
17 | log |
|
log
5
1
5
1
x
x
;
13
17
0
13)
(
17)
(
;
0
13
0
17
13|
17 | |
|
2
2
x
x
x
x
x
x
x
x
13
17
15
;
13
17
0
30)
(2
4
x
x
x
x
x
x
;
13;
15; 13
x
(B)
25(96-9-29)
2 ;
4
log
log
2
5,0
2
x
x
y
;
1
2
4
0
2
4
;
0
2
4
log
0
2
4
2
2
2
5,0
2
x
x
x
x
x
x
x
x
;
0
3
4
0
2
4
;
1
2
4
0
2
4
2
2
2
2
x
x
x
x
x
x
x
x
;
0
)3
1)(
(
0
2)
(2
2)
2
(
x
x
x
x
2
;3 2
2 1;
2
x
(A)
26(96-12-87) )
;
4
4
log
log
2
5,0
2
x
x
y
;
1
4
4
0
4
4
;
0
4
4
log
0
4
4
2
2
2
5,0
2
x
x
x
x
x
x
x
x
25
;
0
1
4
4
0
4
4
;
1
4
4
0
4
4
2
2
2
2
x
x
x
x
x
x
x
x
2 1;
1
2
;0 1
;
0
)1
2
(
0
)1
2
x
x
x x
(E)
27(96-13-28) )
;
4
4
log
log
2
3
2
x
x
y
;
1
4
4
0
4
4
;
0
4
4
log
0
4
4
2
2
2
3
2
x
x
x
x
x
x
x
x
;
0
1
4
4
0
4
4
;
1
4
4
0
4
4
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x x
;
0
)1
2
(
0
)1
2
(B)
28(98-2-34)
;
1
5
1
0
6 ;
log
8
log
log 10
15
log
5
5
p
p
p
p
p
p
5 1;
1
;
5
1
1
0
p
p
p
(D)
29(98-6-25)
0;
5)
log (
log
2
4
,0 2
x
0;
9
;
0
5
4
5
;
0
5
1
5)
log (
2
2
2
2
2
4
x
x
x
x
x
;3
; 3
0;
)3
3)(
(
x
x
x
(B)
30(98-11-39)
1
log 12; 0
log 6
x
x
x
(B)
31(99-10-38)
32;
4
2
log2
x
x
32;
2
2
2
log2
x
x
;
0
0
4)
4)(
(
;
0
0
16
;
0
32
2
2
2
x
x
x
x
x
x
x
x
4;0
x
. Butun yechimlar 1, 2, 3, bularning
yigindisi 6 (E)
32(96-3-88)
;
2)
(
2)
(
9)
log 2(2
)1
log 2( 2
x
x
x
x
1)
;
)9
log 2(
)1
(
log
1
2
2
2
2
x
x
x
;
0
9
2
9
2
1
1
2
x
x
x
x
;
5,4
0
2)
4)(
(
1
;
9
2
0
8
2
1
2
x
x
x
x
x
x
x
x
x 4;1
2)
;
)9
log 2(
)1
(
log
1
2
0
2
2
2
x
x
x
;
0
9
2
9
2
1
1
2
2
x
x
x
x
;
5,4
0
2)
4)(
(
1
2
;
9
2
0
8
2
1
2
2
x
x
x
x
x
x
x
x
bu
tengsizliklar sistemasi yechimga ega emas.
Javob:
x 4;1
(D)
33(96-9-30)
;
)1
(
3,0
log
4)
( 2 5
3,0
log
x
x
x
x
x
x 1 0
bo’lgani uchun x>1 bo’ladi.
;
)1
(
log
)4
5
(
log
1
3,0
2
3,0
x
x
x
x
;
0
4
5
1
4
5
1
2
2
x
x
x
x
x
x
;
0
4
5
0
5
6
1
2
2
x
x
x
x
x
;
0
)1
4)(
(
0
)1
5)(
(
1
x
x
x
x
x
;5
x
(C)
34(96-12-88)
;
2)
(
2)
(
3)
(
2
log 1
5)
2 5
(
2
log 1
x
x
x
x
x
x 3 0
bo’lgani uchun x-2>1 bo’lad.
1)
;)3
log (
)5
5
(
log
0
3
2
1
2
2
1
x
x
x
x
;
0
5
5
3
5
5
3
2
2
x
x
x
x
x
x
;
0
8
6
3
2
x
x
x
0;
2)
4)(
(
3
x
x
x
;4
x
(D)
26
35(96-13-29)
;
2)
(
2)
(
3)
log 2 (
5)
log 2 ( 2 5
x
x
x
x
x
x 3 0
bo’lgani uchun x-2>1 bo’ladi.
1)
;
)3
log (
)5
5
(
log
0
3
2
2
2
x
x
x
x
;
0
5
5
3
5
5
3
2
2
x
x
x
x
x
x
;
0
5
5
0
8
6
3
2
2
x
x
x
x
x
0
2
5
5
2
5
5
0
)2
4)(
(
3
x
x
x
x
x
;4
2
5
5
x
(C)
36(98-11-49)
32
4
log 2
x
x
tengsizlikning har ikki
tomonini 2 asosga ko’ra lgarifmlaymiz.
5;
4)
(log
log 32 ; log
log
2
2
2
4
2
log
2
x
x
x
x
log2 x t
bo’lsin.
0;
5)
1)(
0; (
5
4
5;
4)
(
2
t
t
t
t
t t
;1
5
t
2;
2
;1
log
5
5
2
x
x
2 5 ;2
x
(E)
37(98-4-39)
0
3)
(
log
6
3
1
x
x
;
1) x=6 tengsizlikning yechimi.
2)
4
3
6
;
1
3
3
6
;
0
3)
(
log
0
3
0
6
3
1
x
x
x
x
x
x
x
x
x
;
4;3
x
. Yagona natural yechim 6. (B)
38(98-4-27)
0;
1
log
1
13
7
3
2
2
2
1
2
2
x
x
x
x
x
x
0
13
7
3
2
x x
chunki D<0. Tengsizlikning har
ikki tomonini
13
7
3
2
x x
bo’lib
0
1
log
1
2
2
2
1
2
x
x
x
x
ni hosil qilamiz.
1)
;
0
1
0
1
2
x
x
bundan
x 1
;
2)
x 1
bo’lsa
0
1
2
x
bo’ladi.
Tengsizlikning
2
1
x
har ikki tomonini bo’lib
0
1
log
2
2
1 2
x
x
x
ni hosil qilamiz.
Bu tengsizlik yechimga ega emas chunki
1
1
0
2,
1
2
2
2
x
x
x
. Shunday qilib yagona
musbat yechim
x 1
(D)
39(01-1-24)
t
x
x
x
2
2
2
1; log
log
2
log
bo’lsin.
;
0
1
)1
2)(
; (
0
1
2
;
0
1
2
;
1
2
2
t
t
t
t
t
t
t
t
t
t
2;
1
;1
;
1
0
)1
1)(
2)(
(
t
t
t
t
t
t
1)
2
;0 1
;
0
2
1
;
.0
1
log2
x
x
x
x
x
2)
4;2
;
0
4
2
;
0
2
log
1
2
x
x
x
x
x
Javob:
;2 4
2
;0 1
x
(E)
40(01-1-25)
0;
3
log
|
| log
3
3
x
x
1)
1
;
1
0
3
;
0
log
0
3
log
log
3
3
3
x
x
x
x
x
2)
;
0
1
3
log
2
;
0
0
log
0
3
log
log
3
3
3
3
x
x
x
x
x
x
x
27
;
0
1
3
3
1
;
0
1
2
3
log3
x
x
x
x
x
x
1
;
3
3
1
x
Har ikkovidan
;
3
3
1
x
(E)
41(01-2-28)
;1
2 )
log 2 (3
x
x
1)
;
5,1
0
)1
3)(
(
0
)1
1)(
(
;
3
2
0
3
2
0
1
;
0
2
3
2
3
1
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x
x
x
1
;
x 3
2)
;
0
5,1
0
)1
3)(
(
0
)1
1)(
(
;
0
3
2
0
3
2
0
1
;
0
0
2
3
2
3
1
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Bu tengsizliklar sistemasi yechimga ega emas.
x=-2 yagona butun yechim 1 ta . (D)
42(01-2-75)
;
2
1
3
1
1
0
;1
3
1
log
log 7
3
log
2
1
x
x
x
x
x
6; 1
5
;
6
5
1
0
x
x
x
(D)
43(01-2-80)
2
2
6
9 | 5
lg | 2
x
x
x
y
5;
4;
;1
9 |
9 | 0 |; 2
)1 lg | 2
2
1
2
x
x
x
x
;
5
;
4
0
6
5
;
9 | 0
| 2
lg
0
6
5
2
2
2
x
x
x
x
x
x
x
5
;
4
3
; 2
5
;
4
0
2)
3)(
(
x
x
x
x
x
x
x
;
2,3,4 va 5 butun yechimlar. Bu sonlarning
yig’indisi 14 (E)
44(01-3-24)
0;
7
log6 3
x
;1
7
3
x
18
;
6
3
x
x
. Eng kichik butun yechim -17 (D)
45(01-4-28)
2;
2
5
log
3
1
x
5,2
;2
2 ;
5,2
9;
2
5
0
2
5
x
x
x
x
x
(B)
46(01-6-38)
7;
128)
(2
log
2
1
x
;
2
2
2
2
;
0
2
2
2
2
2
;
0
128
2
2
1
128
2
7
8
7
7
7
7
x
x
x
x
x
x
;7 8
7;
8
x
x
x
. Yagona butun yechim 8 (D)
47(01-6-39)
;
0
25
0
4
;
25
4
;
2
log
2
log
2
2
2
2
2
5
2
2
x
x
x
x
x
x
5;2
;5 2
0 ;
5)
5)(
(
0
2)
2)(
(
x
x
x
x
x
. Bu
oraliqda 8 ta butun son bor. (D)
48(01-7-27)
3;
log
3
| 3;
| log
2
2
x
x
8
8
1
x
. Bu oraliqdagi butun sonlar 2,3,5 va 7.
Bu sonlarning yig’indisi 17 (C)
49(01-7-28)
0;
3)
2log (2
)1
(
log
9
1
3
1
x
x
0;
3)
log (2
)1
(
log
3
1
3
1
x
x
;
5,1
0
1
3
2
1
;
5,1
1
1
3
2
1
;
0
3
2
0
1
0
3
2
1
log
3
1
x
x
x
x
x
x
x
x
x
x
x
5,1 ;2
;
5,1
0
5,1 )
2)(
(
;
5,1
0
5,1
2
x
x
x
x
x
x
x
(A)
50(01-7-35)
4 ;
1
5,0
7)
log3( 2 6
x x
28
;
2
1
2
1
2
7)
3( 2 6
log
x
x
2;
7)
6
(
log
2
3
x x
0 ;
)1
7)(
(
0
2)
8)(
(
;
0
7
6
9
7
6
2
2
x
x
x
x
x
x
x
x
x ;1 2
. Eng katta butun yechim 2 (B)
51(01-9-3)
0;
1,0
log
2 )
log (3
2
2
2
x
0
1,0
log2
bo’lgani uchun
0
2 )
log2 (3
x
bo’lishi kerak.
1;
;1
;1
2
3
x
x
x
(A)
52(01-9-45)
0
2
2 ; ln 1
ln 1
9
5
4
2
x x
bo’lgani uchun tengsizlik yechimga ega emas. (E)
53(01-10-30)
0;
2)
5) log (
6
(
3
2
x
x
x
;
3
0
)1
5)(
(
;
0
2)
(
log
0
5
6
3
2
x
x
x
x
x
x
5;3
x
(E)
54(01-10-31)
0;
2)
2
(
log
2 )
1(
log
2
2
,0
3
x
x
x
Bu
tengsizlik o’rinli bo’lishi uchun quyadigilar
bajarilishi kerak.
1)
;
0
)1
(
0
;
1
2
2
1
2
1
;
0
)2
2
(
log
0
2 )
1(
log
2
2
2
2,0
3
x
x
x
x
x
x
x
x
;1 0
; 1
x
2)
;
0
)1
(
0
;
1
2
2
1
2
1
;
0
)2
2
(
log
0
2 )
1(
log
2
2
2
2,0
3
x
x
x
x
x
x
x
x
Bu tengsizliklar sistemasi yechimga ega emas.
Javob:
;1 0
; 1
x
(D)
55(01-11-28)
5;
2
12
3)
log 12(
x
x
;3 8
;
3
8
;
0
3
5
2
3
x
x
x
x
x
x
. Eng kichik
butun yechim -2 (B).
56(01-11-32)
;
0
1
0
5
0;
3
log
5
2
x
x
x
x
x
;
0
1
5
x
x
x
;1 5
1;0
x
(C)
57(02-1-59)
0;
3log
log
2
2
3
2
x
x
0;
3)
(log
log
2
2
2
x
x
1)
;1
0;
log2
2
x
x
2)
8;
3;
log2
x
x
;8
1
x
(E)
58(02-2-26)
;)
3
log (2
)1
(2
log
2
5,0
x
x
;
0
3
2
0
1
2
3
2
1
2
1
;)
3
log (2
1
2
1
log
2
2
x
x
x
x
x
x
;
3
2
2
1
0
1
2
3
6
2
4
1
;
2
3
1
2
0
3
2
1
2
1
2
x
x
x
x
x
x
x
x
x
x
;
3
2
2
1
0
)1
1)(6
(2
2
x
x
x
x
x
;
3
2
2
1
0
3
1
2
1
2
1
x
x
x
x
x
2
3; 1
1
x
(A)
59(02-3-40)
;1
6
log 2;
2
3
3
2
log
6
6
x
x
;
0
2
3
3
2
0
2
3
4
6
3
2
;
0
2
3
3
2
0
2
2
3
3
2
;
0
2
3
3
2
2
2
3
3
2
x
x
x
x
x
x
x
x
x
x
x
x
x
;
0
3
2
2
3
0
3
2
4
7
;
0
2
3
3
2
0
2
3
7
4
x
x
x
x
x
x
x
x
29
4 ;
7
5,1
;
x
. Eng kichik musbat butun
yechim 2 (A)
60(02-4-42)
10
1
;1
;1 lg
lg
x
x
x
. Eng
kichik butun yechim 1 (D)
61(02-4-43)
5,0 ;
)1
log16 (3
x
4;
1
3
x
x 1
. Eng kichik butun yechim 2 (E)
62(02-5-26)
3 ;
2
3)
log (
2)
2log (
8
8
x
x
3 ;
2
3)
log (
2)
log (
8
2
8
x
x
3
2
4
3
2)
(
;
0
3
0
2
3
2
3
2)
(
log
2
2
8
x
x
x
x
x
x
x
x
;
3
0
3
16
8
;
3
0
4
3
4
4
2
2
x
x
x
x
x
x
x
x
;4
4;3
;
3
0
3)
4) (
(
2
x
x
x
x
(E)
63(02-6-37)
0;
)1
(
log
)
(5
log
2
4
2
2
5
x
x
x
Bu tengsizlik
o’rinli bo’lishi uchun quyadigilar bajarilishi kerak.
1)
;
0
0
4
;
1
1
1
5
;
0
)1
(
log
0
)
5(
log
2
4
2
2
4
2
2
4
2
2
5
x
x
x
x
x
x
x
x
x
bundan
x
2)
;
0
0
5
0
4
;
1
1
0
5
1
5
;
0
)1
(
log
0
)
5(
log
2
4
2
2
2
4
2
2
2
4
2
2
5
x
x
x
x
x
x
x
x
x
x
x
;0 2
2 0;
;
0
0
5)
5)(
(
0
2)
2)(
(
x
x
x
x
x
x
. Butun
yechimlar -1 va 1. 2 ta (C)
64(02-6-38)
0;
)3
7) log (
8
(
2
5
2
x
x
x
1)
2;
;1
3
0;
)3
log (
2
2
5
x
x
x
2)
;
0
3
1
3
0
)1
7)(
(
;
0
3
0
)3
(
log
0
7
8
2
2
2
2
3
2
x
x
x
x
x
x
x
x
;2 7
0;
2)
2)(
(
0
)1
7)(
(
x
x
x
x
x
. Har ikkovidan
7;2
x 2
(B)
65(02-9-35)
;)
lg(27
2
2)
lg(
x
x
;2
)
lg(27
2)
lg(
x
x
;
100
)
2)(27
(
27
2
;
2
)
2)(27
(
lg
0
27
0
2
x
x
x
x
x
x
x
x
;
0
7)
22)(
(
27
2
;
0
154
29
27
2
2
x
x
x
x
x
x
x
x
22;27
7;2
x
. Bu oraliqqa 8 ta butun son
tegishli. (B)
66(02-10-72)
0;
1
)
log
(2
2
2
x
x
1)
1
;
0
1
;
0
0
1
2
x
x
x
x
x
2)
;
0
0
)1
1)(
(
4
;
0
0
1
2
log
;
0
0
1
0
log
2
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x ;1 4
. Har ikkovidan
x ;1 4
(A)
67(02-11-35)
t
x
x
x
3
3
3
log
;1
log
2
2log
bo’lsin.
;
2
0
2)
2)(
0; (
2
2
0;
1
2
2
;1
2
2
t
t
t
t
t
t
t
t
t
9;
9
1
2;
log
2
2;
2
3
x
x
t
Bu
oraliqdagi tub sonlar 2,3,5 va 7. Bularning
yig’indisi 17 (E)
30
68(02-11-36)
0;
3
log
1
4)
log (2
5
3
1
x
x
x
1)
5
2;
5
;
1
0
0
4
2
0
5
x
x
x
x
x
x
x
2)
;
0
5
2
0
3
log
1
4)
(2
log
3
1
x
x
x
x
;
5
2
0
log
4)
log (2
3
3
x
x
x
x
;
5
2
0
4
2
4
;
5
2
1
4
2
;
5
2
0
4
2
log3
x
x
x
x
x
x
x
x
x
x
x
x
;2 4
;
5
2
0
4)
2)(
(
x
x
x
x
x
. Har ikkovidan
5
4;2
x
. Tengsizlikning 3 ta butun yechimi
bor. (D)
69(03-1-12)
4)
4)ln(
(
5 5
x
x
x
.
4;
5
0;
4
0
5
x
x
x
x
Tengsizlikning aniqlanish
sohasiga tegishli yagona butun son 5 va u
tengsizlikning yechimi emas. Tengsizlikning butun
yechimi mavjud emas. (A)
70(03-1-29)
2;
log 3
x
1)
;3
;
0
3)
3)(
(
1
;
3
1
2
x
x
x
x
x
x
2)
1;0
;
0
3)
3)(
(
1
0
;
3
1
0
2
x
x
x
x
x
x
. Har
ikkovidan
;3
1;0
x
(C)
71(03-2-22)
3 ;
2cos5
3
2
log4
x
;1
3
log4 2
x
;
1
3
;
2
3
2
3
3
;
4
3
2
0
3
2
0
3
x
x
x
x
x
x
x
x
1;3
x
. Butun yechimlar -3,-2,-1,0. 4 ta (B).
72(03-3-35)
;
2
1
2
1
0
2 2
9
6
9
,0 2 log 2
log
x
x
;1
2
9
6
9
0; log
2
9
6
9
log
log
2
2
2
2
,0 2
x
x
x
x
4;
18
6
0; 9
2
2; 9
2
9
6
9
2
2
2
x
x
x
x
x
0;
6
1
3
2
0;
2
9
18
2
x
x
x
x
3
6 ; 2
1
x
(D)
73(03-4-35)
4;
4; 10
10
100
2
lg
lg( 2) 2
x
x
;
2
402
;
2
400
2
;
0
2
4
100
2
x
x
x
x
x
x
402;
2
x
Eng katta butun yechim 401 (B)
74(03-6-60)
;1
3)
(2
log
3
1
x
3
2
1
2
11
;
2
3
3
5
;
3
2
3
10
2
;
0
3
2
3
1
3
2
x
x
x
x
x
x
x
(A)
75(03-7-71)
log 16;
1
log
8
3
2
x
log 16;
)1
3 log 16; log (
1
)1
3 log (
1
2
2
2
2
x
x
;1 15
;1
15
0 ;
1
16
1
x
x
x
x
x
(E)
76(03-8-51)
2;
3)
log (4
x
x
31
1)
0 ;
)1
3)(
(
1
;
3
4
0
3
4
1
;
0
3
4
3
4
1
2
2
x
x
x
x
x
x
x
x
x
x
x
x ;1 3
2)
;
75
,0
0
)1
3)(
(
1
;
3
4
0
3
4
1
;
0
3
4
3
4
1
0
2
2
x
x
x
x
x
x
x
x
x
x
x
x
Bu tengsizliklar sistemasi butun yechimga ega
emas, demak
x ;1 3
.
Butun yechimlar 2 va 3. Bu sonlarning yig’indisi
5. (A)
77(03-9-22)
0;
4
log
2
log
2
2
x
x
4;
log
; 2
4
log
0
4)
2)(log
(log
2
2
2
2
x
x
x
x
16
4
x
. Bu oraliqda 4 ta tub son bor. (C)
78(03-9-23)
0;
2,0
log
2
4)
3
8 | log (
|
3
2
5
x
x
x
8
0;
8
)1
x
x
;.
2)
0;
log 9
4)
3
8; log (
5
2
5
x
x
x
;
0
4
3
9
4
3
;
0
4
3
log 9
4
3
log
2
2
2
5
2
5
x
x
x
x
x
x
x
x
0
)1
4)(
(
;
0
2
61
3
2
61
3
;
0
4
3
0
13
3
2
2
x
x
x
x
x
x
x
x
2
61
;4 3
; 1
2
61
3
x
. Butun
yechimlar -2,5 va 8. 3ta (D)
79(03-9-40)
2 ;
2)
log (
)
(
2
1
x
f x
;
0
2
2
2)
(
log
;
0
2
0
2
2)
(
log
2
1
2
1
x
x
x
x
6;
2 ; 2
6
;
2
4
2
x
x
x
x
x
bu oraliqda 4 ta
butun son mavjud. (D)
80(03-12-24)
4;
2)
log (
2
3
x
;
2
0
)7
11)(
(
;
2
0
9
)2
(
;
0
2
9
)2
(
2
2
2
2
x
x
x
x
x
x
x
;2 11
;7 2
x
. Bu oraliqda 18 ta butun son
mavjud. (D)
81(01-12-57)
21;
ln(3 2 27)
x
e
0 ;
)3
3)(
(
0
4)
4)(
(
;
0
9
0
48
3
;
0
27
3
21
27
3
2
2
2
2
x
x
x
x
x
x
x
x
4;3
;4 3
x
.-4 va 4 butun yechimlar. 2ta (E)
32
