Toshloq tumani

Shuning uchun javob : a) 7.


«)-Ь b)-I_; c)l; d)3; e)-L.


5-misol: Tenglamaning yechimlari ko’paytmasini toping:
l°g* 3 • log_3;c 3 = \og_9x 3.
— — = —-— => log. x • log. 3x = log. 9x =>
log₃ x log₃ 3x log₃ 9x
=> log₃ x • (log₃ 3 + log₃ x) = log₃ 9 + log₃ x =>
Yechish: log₃ x + log² x = log₃ 9 + log₃ x => log² x = log₃ 9 => log² x = 2
=> log₃ x = +V2 => Xj = 3⁴² ,x₂ = T^² =>
Javob :c)l.


Xj • x₂ =3^ ■ У⁴² =3° =1.


II.Ko’pgina logarifmik tengsizliklar quyidagi to’rtta holatdan biriga keltirib yechiladi:
1) 0 < a < 1 bo'Isa, log_af(x) > log_ag(x) tengsizlikni yechish uchun quyidagi sistemani


yechish kerak:


I fix) > 0

2. 
   Agar a > 1, bo’lsa, log_a f(x) > log_a g(x) о |
   
3. 
   Agar 0 < a < 1 bo’lsa, log_a /(*) < log_a g(x) о
   

fix) > g(^x) g(x) > 0 lf(x)>g(x)
Uo) > 0

4) Agar a > l bo’lsa, log_a fix) < log_a g(x) <»

f(x) < g(x) f(x) > 0


2. 1) lg(x + l) < 2 =^> lg(x + l) < IglOO =^>


x + 1 < 100


x + 1 > 0





x <99


x > -1


-1 < x < 99

2. 
   log₂(x-3) + log₂(jc-2)
   

Yechish: Logarifmik funktsiya argumentning musbat qiymatlarida aniqlangan, shuning uchuntengsizlikning chap qismi x-3>0 va x-2>0 dama'noga ega. Demak, bu tengsizlikning aniqlanish sohasi x>3 oraliqdir.
log₂ (x - 3) + log₂ (x - 2) < log₂ 2 =^> log₂(x - 3) • (x - 2) < log₂ 2 =^>
f(x-3)-(x-2)<2^>l < 1 => 3 < x < 4.
[x > 3 =^> x > 3 J

2. 
   log j (x² + 2x - 8) > -4 tengsizlikni yeching.
   

Brinchi kvadrat tengsizlikni yechib -6 < x < 4 ga ega bo’lamiz. 2-kvadrat tengsizlikni yechib x < -4, x > 2 ga ega bo’lamiz. Ikkalasini umumlashtirib - 6 < x < -4, 2 < x < 4 javobga ega bo’lamiz.
5. Tengsizlikni yeching. log ₀₁₅ (x + 5)⁴ > log ₀₁₅ (3x -1)⁴
Yechish: Logarifmlarning asoslari 1 dan kichik bo’lgani uchun potensirlaganga


tengsizlik belgisi qarama-qarshisiga o’zgaradi.
0⁴ <(3j-1)⁴ =^0² <(3j-1)² =>
=^> (x + 5)² - (3j -1)² < 0, x Ф -5 =^>
=^> (x + 5 - 3x +1) • (x + 5 + 3x -1) < 0, л: ^ 5 =>
=^> (6 - 2x) • (4x + 4) < 0, x ^ -5 =>
=^> Xj =-l,x₂ = 3,i ^ -5 => i e (—go;—1) u (3;oo)_?j ф -5 =>
2>ig (-oo;-5)u(-5;-l)u(3;oo)
4x-l

6. 
   Tengsizlikni yeching. ^{1 <}
   

Yechish: (—) ⁴ = 16 bo'lgani uchun :


log j (x² + 2x - 8) > log j 16 =>








Yechish:




л


4x-l 4x + 8


< log 1 =>
Ti


4x-l 4x + 8


> 1, x Ф -2


4x -1
4x + 8


> 0 x, = —; x, = -2
4


+


~T
0


_L
4


4x-l
4x + 8
-9


>1


4x -1


1>0


4x + 8
>0=>4x + 8<0


4x + 8
Javob : x < -2, x e (-oo;-2)


4jc -1 - 4jc - 8
4x + 8


x < -2.


>0


₇! log p(———) > 0
^{7) Л} Злг-1.5


tengsizlikni yeching.


lo _> о

8. 
   ^ S. 2x + 9 tengsizlikni eching.
   

2
4

8. 
   Tengsizlikni yeching. log ₂ (3 - 2x) - log, (3 - 2x) > -
   

8 ^
a)(-go;0,5) b)(-co;l,5;) c)(-4;-l); d)(0;l); e)(-co;0).


log₂(3-2jf) = log^₃ (3 - 2xk³ — log j


1


Yechish:


lOgj


^-log (3-2,)>lo_g (I,-


(3-2xf
4


1


1 (3 - 2xf ,
log! > og1
« (3-2x) -


T¹
Js_y


1


1


4 ^< л4

¹ 2x -3 < 0

3 - 2x 2
3 - 2x > 2
2x <3





2x < 1 2v < 3


(3 - 2v) 2


1
x < —
2 1
=> x < —.
3 2
v < —


2


3 - 2x > 0


Javob : a) (-go; 0,5).

10) Tengsizlikni qanoatlantiradigan butun sonlar nechta?

log₂(4-x)-log₂ 7 < 0 a) 6; b)5; c)8; d)7; e)4.
, [4-x > 0
Yechish: log₂(4-.x)₂7^>i ^>x<4 x>-3