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
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;