30 sondir.
32- rasm.
57
Masalan, log
3
9=2, chunk
i
9=3
2
. Shuningdek,
2
5
7
1
log
3; log 5=1; log 1=0.
8
= −
1- misol.
Hisoblang: log
3
81.
Logarifmning ta’rifiga ko‘ra, 3
4
=81 bo‘lgani uchun log
3
81=4.
Logarifmning xossalari
• asosiy logarifmik ayniyat: agar
a>
0,
а≠
1,
b>
0 bo‘lsa,
log
a
b
a
b
=
tenglik
o‘rinlidir;
• agar
a>
0,
а≠
1 bo‘lsa, log
a
1
=
0
;
log
a
a=
1;
• agar
a>
0,
а≠
1 va
x>
0
, y>
0 bo‘lsa,
log ( ) log
log
a
a
a
xy
x
y
=
+
;
• agar
a>
0,
а≠
1 va
x>
0
, y>
0 bo‘lsa,
log
log
log
a
a
a
x
x
y
y
=
−
;
• agar
a>
0
,
а≠
1
,
x>
0
bo‘lsa
log
log
n
a
a
x
n
x
= ⋅
;
• yangi asosga (bir asosdan boshqa asosga) o‘tish formulasi: agar
a>
0,
а≠
1,
x>
0
, b>
0
, b≠
1 bo‘lsa,
log
log
log
b
a
b
x
x
a
=
;
• agar
a>
0,
а≠
1,
b>
0
, b≠
1
bo‘lsa, log
a
b∙
log
b
a=
1.
log
10
x=
lg
x
va log
e
x=
ln
x
kabi belgilash qabul qilingan. (
e
=2,718281...).
Bunda lg
x
−
x
ning o‘nli logarifmi, ln
x
esa
x
ning natural logarifm deyiladi.
f
(
x
)=log
a
x
funksiya (bu yerda
х
– argument,
a>
0,
а≠
1)
a –
asosli
logarif-
mik funksiya
deyiladi.
Logarifmik funksiyaning xossalari:
• aniqlanish sohasi (0;
+∞
) oraliq;
• qiymatlar sohasi
ℝ
=(
−∞
;
+∞
);
• noli:
x=
1, ya’ni log
a
1=0.
•
a>
1 bo‘lsa, logarifmik funksiya (0;
+∞
) oraliqda o‘suvchi;
• 0
1 bo‘lsa, logarifmik funksiya (0;
+∞
) oraliqda kamayuvchi.
2- misol.
Taqqoslang:
1
2
1
log
3
va 0.
1
2
log 1 0
=
, asos
1
2
a
=
, ya’ni funksiya kamayuvchi
1
0
1
2
< <
va
1
0
1
3
< <
bo‘lganidan
1
1
2
2
1
log
log 1
3
>
bo‘ladi. Demak,
1
2
1
log
0
3
>
ekan.
3- misol.
Funksiyaning aniqlanish sohasini toping:
2
2
5
6
( ) log
1
x
x
f x
x
−
+
=
−
.
Bu logarifmik fun
ks
iyaning aniqlanish sohasi
x
ning
2
5
6
0
1
x
x
x
−
+
>
−
teng-
59
U ho
l
da:
1
lg54 lg
3lg3 3
2
.
lg 72 lg8 2lg3 2
+
=
=
−
Javob
:
1
lg54 lg
3lg3 3
2
.
lg 72 lg8 2lg3 2
+
=
=
−
Eng sodda logarifmik tenglama
log
a
x=b
ko‘rinishdagi tenglamani (
a>
0,
a
≠
1
, b
– haqiqiy son) eng sodda
logarifmik tenglama deyish mumkin. Tenglamaning yagona yechimi:
x=a
b
.
7- misol.
Tenglamani yeching:
3
1
log
2
x
=
.
Logarifm ta’rifiga ko‘ra,
yechimi
1
2
3
3
x
=
=
.
Javob
:
3.
x
=
8- misol.
Tenglamani yeching:
log 16 2
x
=
.
Logarifmning ta’rifiga ko‘ra,
x
2
=
16 va
x>
0,
x
≠
1
Demak, tenglamaning ye
-
chimi
х=
4 ekan.
Javob
:
х=
4.
9- misol.
Tenglamani yeching:
2
2
log (
5 10) 4
x
x
−
+
=
.
Logarifmning ta’rifiga ko‘ra,
2
4
5 10 2
x
x
−
+
=
tenglamani hosil qilamiz.
Kvadrat tenglamani yechib
x
1
=−
1,
x
2
=
6
ildizlarni topamiz. Demak, tenglama
-
ning yechimi {−1; 6} ekan.
Javob
:
x=−
1,
x=
6.
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