logarifmik funksiyalar, tenglamalar va tengsizliklar

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1576158303.doc ) 2 ( 3 log - = x y 2 5 25 1 100 4 04 , 0 - = = = 4 3 4 log ) 4 , 2 1 log 4 - = = x x 2 4 2 1 = = x 4 4 log = x x 3 3 4 4 3 256 1 4 , 4 = = = - - x x , log 1 log ) 6 , log log ) 5 , log log log ) 4 , log log ) ( log ) 3 , 1 log ) 2 , 0 1 log ) 1 x m x x n x y x y x y x xy a a a a n a a a a a a a a a m = = - = + = = = , log 1 log ) 9 , log log …

7 5 2 . 12 log ) 4 ; 9 log ) 3 ; 15 log ) 2 ; 4 log ) 1 ) 306 3 , 1 8 , 0 7 3 . 2 1 ) 4 ; 3 2 ) 2 . 303 ; 16 1 ) 4 - . 2 ) 10 ; 0 ) 8 ; 3 40 log ) 6 8 x log x log 1 a 2 a > 1 2 log log x x a a a a > x log x log 1 a 2 a > b a b a = log ) 2 ( log 3 1 - = x y )2(log 3 1  xy ) 2 ( 3 log - = x y )2( 3 logxy ) 2 ( log 3 1 - = x y a 1 0 1 ) 2 ( log 3 1 - = …

2 ( - î x ) 1 ( log ) 4 2 ( log 3 1 3 1 + > - x x 1 3 1 + > - 1 4 2 0 1 0 4 2 x x x x ï î ï í ì > 5 1 2 x x x ) 5 , 2 ( î x 2 ) 12 ( log ) 2 ( log 3 1 3 1 - ³ - + - x x î í ì > - > - 0 12 0 2 x x î í ì 12 2 x x 2 3 1 3 1 ) 3 1 ( log ) 12 )( 2 ( log - ³ - - x x î í ì - x x x . 2 ) 6 5 ( log ) 6 ; 1 ) 3 , 0 2 ( log ) 5 ; …

= - = - - + + y x xy y x y x y x y x y x y x y x y x ). 5 , 12 ( ) 8 ); 4 , 3 ( ) 6 ); 1 , 2 ( ) 4 ; 3 8 : 6 7 ) 2 ÷ ø ö ç è æ ). 7 , 2 ( ) 4 ); 3 , 3 ( ), 1 , 5 ( ) 2 . 323 . 2 , 3 4 ) 4 ); 3 ; 5 , 0 ( ) 2 - ÷ ø ö ç è æ - î í ì = + - = - 0 1 2 1 log log 2 3 3 x y y x y x y x y x 3 , 3 , 1 log 3 = = = 1 , 2 1 2 1 = …

) 16=24 bo`lgani uchun, 16 ni hosil qilish uchun ikkini to`rtinchi darajaga ko`tarish kerak, demak log216=4. 2) ekanligi ma`lum. shuning uchun log50, 04=-2 misollar: 3. tenglamalarni qanoatlantiruv-chi x larni topamiz. yechish: asosiy logarifmik ayniyatdan foydalanib: 3) 4) , ya`ni larni topamiz. har qanday a>0, b>0, a≠1, b≠1, x>0, y>0 va haqiqiy istalgan n va m sonlar uchun quyidagi tengliklar bajariladi: bu tengliklar ko`rsatkichli funksiya xossalaridan kelib chiqadi. bulardan ba`zilarini isbot qilamiz. logarifmik ayniyatdan foydalanib: ni topamiz. bu tengliklarni hadlab ko`paytirsak yoki bo`lsak hosil bo`ladi. bu tengliklardan logarifm ta`rifiga ko`ra 3) va 4) tengliklar kelib chiqadi. ayniyatning ikkala tomonini n – darajaga oshirsak, hosil bo`lib, bundan ni topamiz. bir asosli logarifmdan boshqa asosli logarifmga o`tish formulasi 8) ni xususiy holda 9) ni isbotlash uchun quyidagicha amal qilamiz: hosil bo`lgan x=ab ifodaning ikkala tomonidan b asosga ko`ra logarifm topamiz: chap tomonga b ning qiymatini qo`yib, 8) formulani hosil qilamiz. agar bu formuladan …

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1576158303.doc ) 2 ( 3 log - = x y 2 5 25 1 100 4 04 , 0 - = = = 4 3 4 log ) 4 , 2 1 log 4 - = = x x 2 4 2 1 = = x 4 4 log = x x 3 3 4 4 3 256 1 4 , 4 = = = - - x x , log 1 log ) 6 , log log ) 5 , log log log ) 4 , log log ) ( log ) 3 , 1 log ) 2 , 0 1 log ) 1 x m x x n x y x y x y x xy a a …

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