Toshloq
tumani
Bu ildizlar (1) tenglamaning ildizlari bo‘lishi yoki bo‘lmasligini tekshiramiz.
:
1
x
.
3
2
1
4
log
2
log
)
3
1
(
log
)
1
1
(
log
2
2
2
2
Demak:
1
x
tenglamaning ildizi.
:
5
x
bu holda
1
x
va
3
x
Lar manfiy sonlar bo‘ladi va shuning uchun (1)
tenglamaning chap qismi ma`noga ega bo‘lmaydi. Demak:
5
x
(1) tenglamaning
ildizi emas.Javob:
1
x
.
2-misol:
).
3
lg(
lg
)
12
4
2
lg(
2
x
x
x
x
Yechish:
.
4
,
3
0
12
7
)
3
(
12
4
2
)
3
(
lg
)
12
4
2
lg(
2
1
2
2
2
x
x
x
x
x
x
x
x
x
x
x
x
Tekshirishlar
x
ning ikala qiymati ham dastlabki tenglamaning ildizi ekanligini
ko‘rsatadi.
Javob:
3
1
x
,
.
4
2
x
3-misol:
).
8
5
(
log
)
4
3
(
log
7
7
x
x
Yechish:
Logarifm ishorasi ostidagi ifodalarni tenglaymiz:
.
2
4
2
4
8
5
3
8
5
4
3
x
x
x
x
x
x
2
x
qiymatni berilgan tenglamaga olib borib qo‘ysak, tenglamaning chap va o‘ng
qismlari ma`noga ega bo‘lmaydi.
Javob:, Tenglama ildizga ega emas.
I.
Ko‘pchilik logarifmik tenglamalar quyidagi uchta qoidadan biri yordamida
yechiladi.
1)
.
)
(
)
(
log
b
a
a
x
f
b
x
f
2)
0
)
(
0
)
(
)
(
)
(
)
(
log
)
(
log
x
g
x
f
x
g
x
f
x
g
x
f
a
a
3)
.
1
)
(
,
0
)
(
)
(
)
(
)
(
log
)
(
x
f
x
f
x
g
x
f
b
x
g
b
x
f
1.
1-misol:
;
16
1
e)
;
4
1
-
d)
;
8
1
c)
;
8
1
-
b)
;
16
1
-
)
0
)
1
(
log
log
log
2
2
18
a
x
Yechish:
.
4
1
2
1
2
)
1
(
log
1
)
1
(
log
log
0
)
1
(
log
log
log
2
2
2
2
2
2
18
x
x
x
x
x
Javob:
.
4
1
-
)
d
2-misol:
4.
e)
1;
-
va
2
d)
2;
c)
3;
b)
3;
va
2
a)
.
2
)
8
3
(
log
3
x
x
Toshloq tumani
Yechish:
2.
c)
:
Javob
2.
x
,
9
3
b)
yo`q;
ildiz
1
3
a)
.
9
,
1
0
9
8
t
kiritsak,
belgilash
3
0
9
3
8
)
3
(
3
9
8
3
3
8
3
x
x
2
1
2
x
2
2
t
t
t
t
x
x
x
x
x
x
3-misol:
1.
-
3;
e)
1;
-
2;
d)
1;
-
c)
2;
b)
3;
a)
;
1
1
log
2
x
Yechish:
.
1
x
-2
1
-
x
b)
3
x
2
1
-
x
a)
.
2
1
2
1
2
1
1
x
x
Javob: e) 3: -1.
4-misol:
9.
e)
4;
d)
8;
c)
6;
b)
7;
a)
0
)
1
lg(
3
)
169
lg(
3
x
x
Yechish:
7.
a)
:
javob
uchun
Shuning
irmaydi.
qanoatlant
1)ni
lg(x
ildiz
8
x
.
8
,
7
0
56
0
168
3
3
1
3
3
169
)
1
(
169
10
)
1
(
169
0
)
1
(
169
lg
2
2
1
2
2
2
3
3
3
3
0
3
3
3
3
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
5-misol: Tenglamaning yechimlari ko‘paytmasini toping:
.
3
log
3
log
3
log
9
3
x
x
x
.
2
1
e)
3;
d)
1;
c)
;
3
1
-
b)
;
3
1
)
a
Yechish:
1.
c)
:
Javob
.
1
3
3
3
3
,
3
2
log
2
log
9
log
log
log
9
log
log
log
log
9
log
)
log
3
(log
log
9
log
3
log
log
9
log
1
3
log
1
log
1
0
2
2
2
1
2
2
2
1
3
2
3
3
2
3
3
3
2
3
3
3
3
3
3
3
3
3
3
3
3
3
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
II.
Ko‘pgina logarifmik tengsizliklar quyidagi to‘rtta holatdan biriga
keltirib
yechiladi:
g(x)
log
f(x)
log
bo`lsa,
1
a
0
)
1
a
a
tengsizlikni yechish uchun quyidagi sistemani
yechish kerak:
0
)
(
)
(
)
(
x
f
x
g
x
f
2) Agar
1
a
, bo‘lsa,
)
(
log
)
(
log
x
g
x
f
a
a
0
)
(
)
(
)
(
x
g
x
g
x
f
3) Agar
1
0
a
bo‘lsa,
)
(
log
)
(
log
x
g
x
f
a
a
0
)
(
)
(
)
(
x
g
x
g
x
f
Toshloq tumani
4) Agar
1
a
bo‘lsa,
)
(
log
)
(
log
x
g
x
f
a
a
0
)
(
)
(
)
(
x
f
x
g
x
f
2.
1)
100
lg
)
1
lg(
2
)
1
lg(
x
x
0
1
100
1
x
x
99
1
1
99
x
x
x
2)
1
)
2
(
log
)
3
(
log
2
2
x
x
tengsizlikni yeching.
Yechish: Logarifmik funktsiya argumentning musbat qiymatlarida
aniqlangan,
shuning uchun tengsizlikning chap qismi
0
2
-
x
va
0
3
x
da ma`noga ega. Demak,
bu tengsizlikning aniqlanish sohasi x>3 oraliqdir.
2
log
)
2
(
)
3
(
log
2
log
)
2
(
log
)
3
(
log
2
2
2
2
2
x
x
x
x
.
4
3
3
x
3
4
1
2
)
2
(
)
3
(
x
x
x
x
x
3)
4
)
8
2
(
log
2
2
1
x
x
tengsizlikni yeching.
Yechish:
:
uchun
bo`lgani
16
)
2
1
(
4
16
log
)
8
2
(
log
2
1
2
2
1
x
x
0
8
2
16
8
2
2
2
x
x
x
x
Brinchi kvadrat tengsizlikni yechib
4
6
x
ga ega bo‘lamiz. 2-kvadrat
tengsizlikni yechib
2
x
,
4
x
ga ega bo‘lamiz. Ikkalasini umumlashtirib
4
x
2
,
4
6
x
javobga ega bo‘lamiz.
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