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a > 0 va a 6= 1 bo’lsa, log
√ a 3 √ a ifodaning qiymatini toping. A) 2 3 B) 3 2 C) 3 D) 6 32. (02-3-33) Hisoblang. 1 log 2 4 + 1 log 4 4 + 1 log 8 4 + 1 log 16 4 + + 1 log 32 4 + 1 log 64 4 + 1 log 128 4 A) 14 B) 16 C) 7 D) 32 33. (02-5-24) Hisoblang. log −1 3 3 r 3 q 3 √ 3 A) 27 B) −27 C) 1 27 D) 3 Yechish: Logarifmning 5-xossasi va ildizning xos- sasidan foydalansak log −1 3 3 r 3 q 3 √ 3 = 1 log 3 3 1 27 = 1 1 27 log 3 3 = 27 bo’ladi. Javob: 27 (A). 34. (03-5-39) Hisoblang. y = log 5 log 5 5 s 5 r 5 q 5 √ 5 A) −4 B) 1 5 C) − 1 4 D) 4 35. (02-10-73) Hisoblang. Ã log 6 27 + 2 log 6 2 log 6 3 √ 0, 25 + log 6 1 3 ! 3 A) −27 B) 27 C) −8 D) 8 log 6 27 36. (02-12-48) Hisoblang. 3 lg 2 + 3 lg 5 lg 1300 − lg 0, 13 A) 0, 8 B) 0, 6 C) 0, 7 D) 0, 75 37. (03-2-20) Hisoblang. 1 + 2 log 3 2 (1 + log 3 2) 2 + log 2 6 2 A) 2 B) 0, 5 C) 1 D) 1 4 38. (03-3-33) Hisoblang. log 8 5 2 log 25 32 A) 1 2 B) 1 3 C) 5 3 D) 2 39. (03-4-32) ln(3 log 3 0,64 +8 log 8 0,36 ) ning qiymati −11 dan qancha ko’p? A) 10 B) 9 C) 11 D) 12 40. (03-4-33) Hisoblang. 2 log 4 8 − 3 log 8 4 + log 2 32 + 18 A) 22 B) 24 C) 26 D) 20 41. (98-5-29) Agar log 3 5 = a, log 3 2 = b bo’lsa, log 6 45 ni a va b orqali ifodalang. A) b + 2 a + 2 B) 2 + a 1 + b C) a 1 + b D) b 1 + a Yechish: log 3 5 = a, log 3 2 = b berilgan. Bir asosdan boshqa asosga o’tish formulasi 7-ga ko’ra log 6 45 = log 3 (9 · 5) log 3 (2 · 3) = log 3 9 + log 3 5 log 3 2 + log 3 3 = 2 + a b + 1 ekanini hosil qilamiz. Javob: 2 + a b + 1 (B). 42. (96-10-36) Agar log 4 125 = a bo’lsa, lg 64 ni a orqali ifodalang. A) 3 2 a + 4 B) 2 3 a + 6 C) 18 2a + 3 D) 6 3a + 2 43. (96-9-28) Agar a = log 50 40 bo’lsa, log 5 2 ni a orqali ifodalang. A) 3a − 1 2 − a B) a − 3 1 − 2a C) a − 3 2a − 1 D) 1 − 2a a − 3 44. (96-3-86) Agar a = log 98 56 bo’lsa, log 7 2 ni a orqali ifodalang. A) 3 − a 2a − 1 B) 2a − 1 3 − a C) a − 3 2a − 1 D) 1 − 2a 3 − a 128 45. (00-1-38) Agar a = log 12 2 bo’lsa, log 6 16 ning qiymatini toping. A) 4a 1 + a B) 2a 1 − a C) 4a 1 − a D) 3a 1 + a 46. (00-6-32) Agar log 0,5 27 = a bo’lsa, log √ 3 6 √ 1, 5 ning qiymatini toping. A) 1 3 + a −1 B) a 2 − 1 C) 3 + a −1 D) 1 + a −3 47. (00-10-34) Agar a = log 2 3 bo’lsa, log 8 0, 75 ni a orqali ifodalang. A) 1 3 (a − 1) B) 1 3 (a + 1) C) 1 3 (a − 2) D) 1 3 (a + 2) 48. (00-10-66) Agar log a 27 = b bo’lsa, log √ 3 6 √ a ni toping. A) 1 b B) 2 b C) − b 2 D) 2b 49. (99-10-35) Agar log 2 a = 2 va log 3 b = 2 bo’lsa, log 6 ab ning qiymatini toping. A) −2 B) 3 C) −3 D) 2 50. (00-8-38) Agar lg 5 = a va lg 3 = b bo’lsa, log 30 8 ni a va b orqali ifodalang. A) 3 − 3a 1 + b B) 3(1 − b) 1 + a C) 3(a − b) a + b D) b − 1 a + 1 51. (01-8-31) Agar log 0,2 27 = a bo’lsa, log √ 3 6 √ 1, 8 ni a orqali ifodalang. A) a 2 − 2 3 B) a −1 + 1, 5 C) a −3 + 2 D) a −1 + 2 3 52. (02-8-12) Agar 7 log 5 b = 4 bo’lsa, b log 5 √ 7 ni hisoblang. A) 2 B) 3 C) 1 D) 4 53. (02-8-13) Agar lg 2 = a va lg 3 = b bo’lsa, log 9 20 ni a va b orqali ifodalang. A) 1 + a 2b B) 1 − a 2b C) b 1 + 2a D) b 1 − 2a Yechish: Boshqa asosga o’tish formulasi 7-ga ko’ra log 9 20 = lg 20 lg 9 = lg 10 + lg 2 lg 3 2 = 1 + a 2b ekanini hosil qilamiz. Javob: 1 + a 2b (A). 54. (02-9-38) Agar log b a ( a 2 b ) = − 1 2 bo’lsa, log a 2 b (ab) ni hisoblang. A) − 1 4 B) −1 C) 1 D) 0, 8 55. (02-10-27) lg 2 = a va log 2 7 = b bo’lsa, lg 56 ni a va b orqali ifodalang. A) 3a+ab B) 2a+3b C) 3a+2b D) 2a + 5b 3 56. (03-4-37) Agar log a 8 = 3 va log b 243 = 5 bo’lsa, ab ning qiymatini toping. A) 4 B) 5 C) 6 D) 8 57. (03-7-67) Agar lg 5 = a va lg 3 = b bo’lsa, log 30 8 ni a va b orqali ifodalang. A) a 2a + 3b B) b − 3 1 − 2a C) 3a − 3 b + 2 D) 3(1 − a) 1 + b 58. (03-8-43) Agar a = log 5 4 va b = log 5 3 bo’lsa, log 25 12 ni a va b orqali ifodalang. A) a + b 2 B) a − b 4 C) ab 2 D) a 2 + b 4 12.2 Logarifmik tenglamalar O’zgaruvchisi logarifm belgisi ostida kelgan tenglamalar logarifmik tenglamalar deyiladi. Masalan, log 2 x = 3, log x 2 = 1, log 3 (x 2 − 5x + 3) = 0. Eng sodda logarifmik tenglamaning ko’rinishi log a x = b bo’lib, uning yechimi x = a b . Agar log a f (x) = log a g(x) (1) tenglamada f (x) > 0, g(x) > 0 shart bajarilganda, u f (x) = g(x) (2) tenglamaga teng kuchli bo’ladi. (1) tenglamadan (2) tenglamaga o’tganda chet ildizlar paydo bo’lishi mumkin. Chet ildizlarni aniqlash uchun, ildizlarni dastlabki tengla- maga qo’yib tekshirib ko’riladi. Quyidagi teng kuchlilik- larni keltiramiz. 1. log a f (x) = b ⇐⇒ f (x) = a b . 2. log a f (x) = log a g(x) ⇐⇒ f (x) = g(x), f (x) > 0, g(x) > 0. 3. log f (x) g(x) = b ⇐⇒ ½ [f (x)] b = g(x), f (x) > 0, f (x) 6= 1. A. log a f (x) = b, log a f (x) = log a g(x) ko’rinishdagi tenglamalar 1. log 5 x = 2 tenglamani yeching. A) 10 B) 25 C) √ 5 D) 32 Yechish: 1-xossaga ko’ra x = 5 2 = 25. Javob: 25 (B). 2. lg x = −1 tenglamani yeching. A) 10 B) 0, 1 C) √ 10 D) −1 3. ln x = ln(8 − x) tenglamani yeching. A) e 2 B) 0, 8 C) √ 8 D) 4 4. log 2 x 2 = 4 tenglamani yeching. A) 4 B) ±2 C) ±4 D) 16 5. log 2 log 3 x = 0 tenglamani yeching. A) 8 B) ±3 C) 3 D) 9 6. lg(x − 4) = lg(4 − x) tenglama nechta ildizga ega. A) 1 B) 2 C) 0 D) 4 129 7. (00-7-33) a ning qanday qiymatlarida lg x + lg(x − 6) = lg(−a) tenglama bitta ildizga ega bo’ladi? A) 9 B) a ∈ (−∞; 0) C) 7 D) 6 Yechish: Berilgan tenglamaning aniqlanish so- hasi x > 6 to’plamdan iborat. log a b + log a c = log a bc formula yordamida tenglamaning chap qis- mini almashtiramiz. lg x(x − 6) = lg(−a). Uni potenserlaymiz. x(x − 6) = −a (a < 0). Hosil bo’lgan tenglamani yechamiz. x 2 − 6x + a = 0; D = 36 − 4a = 4(9 − a). Bu tenglama a ≤ 9 da yechimga ega va uning ildizlari quyidagilar x 1,2 = 6 ± 2 √ 9 − a 2 = 3 ± √ 9 − a x 1 = 3 − √ 9 − a ≤ 3 bo’lgani uchun u beril- gan tenglamaning aniqlanish sohasiga kirmaydi. Demak, u chet ildiz. x 2 = 3 + √ 9 − a berilgan tenglamaning ildizi bo’lishi uchun x 2 > 6 bo’lishi kerak. Bu tengsizlikdan a ni topamiz. √ 9 − a > 3 ⇐⇒ 9 − a > 9 ⇐⇒ a < 0. Shunday qilib, a ∈ (−∞; 0) da berilgan tenglama bitta ildizga ega. Javob: a ∈ (−∞; 0) (B). 8. (98-9-34) Tenglamani yeching. lg(x 2 + 2x − 3) = lg(x − 3) A) 0 B) −1 C) 0; −1 D) ∅ 9. (99-6-26) Tenglamani yeching. log 18 log 2 log 2 ³ − 1 x ´ = 0 A) − 1 16 B) − 1 8 C) 1 8 D) − 1 4 10. (99-6-50) Tenglamani yeching. log 1 5 log 5 √ 5x = 0 A) −5 B) 1 C) 0 D) 5 11. (00-2-22) Agar ½ 3 x · 2 y = 972, log √ 3 (x − y) = 2 bo’lsa, xy ning qiymatini toping. A) 14 B) 12 C) 10 D) 8 12. (01-3-26) Tenglama ildizlari yig’indisini toping. lg ³ 3 q x2−4x x−3 + 1 ´ = 1 A) 10 B) 2 C) 8 D) 25 13. (01-7-25) Tenglamani yeching lg(3 + 2 lg(1 + x)) = 0 A) 0 B) 1 C) −15 D) −0, 9 Yechish: Berilgan tenglamani lg(3 + 2 lg(1 + x)) = lg 1 shaklda yozamiz. Bu yerdan 3+2 lg(1+x) = 1 ni, undan esa lg(1+x) = −1 ni olamiz. Ta’rifga ko’ra 1 + x = 10 −1 ⇐⇒ x = −0, 9. Bevosita tekshirish ko’rsatadiki, x = −0, 9 tenglamani qanoatlanti- radi. Javob: −0, 9 (D). 14. (01-7-26) Tenglamani yeching log 2 |x − 1| = 1 A) 3 B) 2 C) −1 D) 3; −1 15. (01-9-41) Ushbu lg(5x − 2) = lg(2 − 5x) tenglamaning aniqlanish sohasini toping. A) (0, 4; ∞) B) ∅ C) (−∞; 0, 4) D) {2,5} 16. (02-3-35) Tenglama ildizlari ayirmasining moduli nechaga teng? log 3 (4 · 3 x − 1) = 2x + 1 A) 1 B) 2 C) 3 D) 0 17. (02-10-69) Tenglamani yeching. log 2 (2 2x + 16 x ) = 2 log 4 12 A) log 4 3 B) log 2 3 C) 2 D) log 4 6 18. (02-10-71) Agar ½ log 2 (x − y) = 1 2 x · 3 y+1 = 72 bo’lsa, x va y ning o’rta proporsional qiymatini toping. A) √ 3 B) 2 C) √ 2 D) 2 √ 2 B. a log a f (x) = g(x) ko’rinishdagi tenglamalar 19. (96-6-55) Tenglamaning ildizini toping. 3 2 log 3 x = 16 A) 3 B) −4 C) 4 D) ±4 Yechish: Tenglama x > 0 da aniqlanish. Loga- rifmning 5-xossasidan foydalanib, uni 3 log 3 x 2 = 16 shaklda yozamiz. Asosiy logarifmik ayniyatdan x 2 = 16 ni olamiz. Bu yerdan x 1 = −4, x 2 = 4 kelib chiqadi. x 1 = −4 tenglamaning aniqlanish sohasida yotmaydi. x = 4 tenglamani qanoat- lantiradi. Javob: 4 (C). 130 20. (97-2-55) Tenglamaning ildizi 20 dan qancha katta? 4 log 4 (x−5) = 19 A) 6 B) 2 C) 4 D) 3 21. (97-8-40) Tenglamani yeching. 4 2 log 4 x = 25 A) 5 B) ±5 C) −5 D) 10 22. Tenglamani yeching. 2 log 4 x = 1 4 A) 1 B) 2 C) 0, 5 D) 0, 25 23. (01-5-12) Tenglamani yeching x log x (x 2 −1) = 3 A) 2 B) 1 C) 3 D) 4 C. 3-4 va 5-xossalariga oid tenglamalar 24. (97-12-54) Tenglamaning ildizi 8 dan qanchaga kam? log 2 (x + 2) + log 2 (x + 3) = 1 A) 7 B) 9 C) 10 D) 6 Yechish: Tenglamaning aniqlanish sohasi x > −2 to’plam. 3-xossadan foydalansak, tenglamani log 2 (x+2)(x+3) = log 2 2 shaklda yozish mumkin. Bu yerdan (x + 2)(x + 3) = 2 ⇐⇒ x 2 + 5x + 4 = 0 kelib chiqadi. Bu kvadrat tenglamaning ildizlari x 1 = −4, x 2 = −1 dir. x 1 = −4 tenglamaning aniqlanish sohasida yotmaydi. x = −1 tenglamani qanoatlantiradi. 8 − (−1) = 9. Javob: 9 (B). 25. (00-3-38) Tenglamani yeching lg ³ 1 2 + x ´ = lg 1 2 − lg x A) 2 B) 1 2 C) 1 D) −1 26. (99-3-20) Tenglamani yeching lg √ x − 5 + lg √ 2x − 3 + 1 = lg 30 A) 1 2 B) 6 C) 1 2 ; 6 D) 1 2 ; 8 27. (02-12-50) Agar lg(x 2 + y 2 ) = 2, lg 2 + lg xy = lg 96 va x > 0 bo’lsa, x+y yig’indining qiymatini toping. A) 12 B) 14 C) 16 D) 18 28. (03-7-21) Tenglamani yeching. log √ 5 (4 x − 6) − log √ 5 (2 x − 2) = 2 A) 3 2 B) 5 4 C) 2 D) 2,5 29. (99-6-28) Tenglamani yeching. log 2 (54 − x 3 ) = 3 log 2 x A) −3 B) 2 C) 1 D) 3 Yechish: 5-xossadan foydalansak, tenglamani log 2 (54 − x 3 ) = log 2 x 3 shaklda yozish mumkin. Bu yerdan 54 − x 3 = x 3 ⇐⇒ 27 = x 3 ⇐⇒ x = 3 kelib chiqadi. x = 3 tenglamani qanoatlantiradi. Javob: 3 (D). 30. (00-2-24) log 5 x = 2 log 5 3 + 4 log 25 7 bo’lsa, x ni toping. A) 441 B) 125 C) 256 D) 400 31. (00-3-28) Tenglamani yeching. ³ 4 9 ´ x · ³ 27 8 ´ x−1 = lg 4 lg 8 A) 3 B) 4 C) 2 D) 1 32. (00-8-15) Tenglamani yeching. log 2 (9 x−1 + 7) = 2 log 2 (3 x−1 + 1) A) 2 B) 1 C) 3 D) 4 33. (01-5-11) Tenglamani yeching log a x − log a 2 x + log a 4 x = 3 4 A) a B) a 2 C) a 4 D) 2 34. (03-4-34) Agar log 4 (2 − x) 2 (3 − x) 3 = −3 log 4 |3 − x| bo’lsa, x − 27 ni hisoblang. A) −25 B) −29 C) −26 D) −24 35. (03-11-13) Tenglamani yeching. 7 (2x 2 −5x−9)/2 = ( √ 2) 3 log 2 7 A) −1, 5; 1 B) 1, 5 C) −2, 5; 4 D) −1, 5; 4 D. 6-7 va 8-xossalarga oid tenglamalar 36. (99-6-55) Tenglamani yeching. log √ 2 x + 2 log x 2 = 4 A) 2 B) 1 C) 3 D) 4 Yechish: 6-8-xossalardan foydalanib, tenglamani 1 1 2 log 2 x + 2log 2 x = 4 ⇐⇒ 4log 2 x = 4 shaklda yozamiz. Bu yerdan log 2 x = 1 kelib chiqadi. Logarifm ta’rifidan x = 2 1 ni olamiz. Javob: 2 (A). 131 37. (98-11-45) Tenglama ildizlari ko’paytmasini top- ing. log x 2 log 2x 2 = log 4x 2 A) 1 B) 1 √ 2 C) − 1 √ 2 D) 1 2 38. (99-3-21) Tenglamani yeching. log 4 (x + 12) · log x 2 = 1 A) 4 B) −3 C) 2 D) 4;2 39. (02-3-36) Tenglama ildizlari ko’paytmasini toping. log x 2 + log 4x 4 = 1 A) 2 B) 4 C) 1 D) 8 E. 11-xossa yoki logarifmlash yordamida yechiladigan tenglamalar 40. (97-6-59) Tenglamani yeching. x lg 9 + 9 lg x = 6 A) 1 B) 10 C) √ 10 D) 2 Yechish: 11-xossadan foydalanib, tenglamani 9 lg x + 9 lg x = 6 ⇐⇒ 2 · 9 lg x = 6 ⇐⇒ 9 lg x = 3 shaklda yozamiz. Bu tenglikning ikkala qismini 3 asosga ko’ra logarifmlaymiz log 3 9 lg x = log 3 3 ⇐⇒ lg x · 2 = 1 ⇐⇒ lg x = 1 2 . Logarifm ta’rifidan x = 10 1/2 ni olamiz. Javob: √ 10 (C). 41. (00-3-39) Tenglama ildizlari ko’paytmasini toping. x lg x−1 = 100 A) 10 B) 20 C) 100 D) 1 42. (01-2-73) Tenglama ildizlari ko’paytmasini toping. x (lg x+5)/3 = 10 5+lg x A) 100 B) 10 C) 1 D) 0,01 43. (01-9-9) Tenglama ildizlarining o’rta proporsional qiymatini toping. x 3−log 3 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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