Chuyên Đề 2: ph ng trình logaritươ I. Phương trình, bất phương trình mũ : 1/ Đưa về cùng một cơ số hoặc hai cơ số: 2 1 8 1 3 1 2 3 4 1/ 3 0,2 1/ 2 4 ; 2 / 3 3 3 3 750; 3 /5 .8 500 (5.2 ) 1 3; 2 x x x x x x x x x x x x x log − − + − + − − − − = + − + = = ⇔ = ⇒ = 1 1 1 1 4 / ( 5 2) ( 5 2) 1 ( 2; 1) (1; ) 1 x x x x x x x − − + − + ≥ − ⇔ − ≥ − ⇒ ∈ − − ∪ +∞ + ( ) 1 2 1 2 9/4 5 / 9 9 9 4 4 4 9 .91 4 .21 9 / 4 21/ 91 log (21/ 91) x x x x x x x x x x + + + + + + < + + ⇔ < ⇔ < ⇔ < 2 2 3 1 2 6 / 2 .4 256; 7 / 2 .5 0,01; 8 / 2 . 3 216; 9 / (3 3 3) (1/ 81) ; 10 / 2 .3 .5 12 x x x x x x x x x x x+ − − = = = = = 2 2 4 2 2 2 3 1 1/ 2 1 1/(3 1) 1 3 11/ 2 5 12 / 8 36.3 ; 13 /1 5 25; 14 / 2 2 ; 15 / ( 10 3) ( 10 3) x x x x x x x x x x x x x − − − − + − + − + − + = = < < ≥ + < − 2/ Đặt ẩn phụ: 2 2 2 2 2 2 1 2 1/ (7 4 3) 3(2 3) 2 0( 3 / 2 0); 2 / (3 5) (3 5) 2 0 x x x x x x x x t t − − + − + − − + = − + = + + − − ≤ 3 3( 1) 1 2 4 4 ( 1/ 2 0); 3 / 2 6.2 1/ 2 12 / 2 1( 2 2 ); 4 / 3 8.3 9.9 0 x x x x x x x x x x t t t − − + + + + − ≤ − − + = = − − − > ( chia 2 vế cho 2 3 x ); 2 2 2 1 2 3 2 1 5/ 4 5.2 6 0;6/4 7.4 2 0 x x x x cosx cosx+ − − + − + + − − = − − = ; 2 2 2 2 1 4 1 7 / 27 6.64 6.36 11.48 ; 8 / 2 2 2 ; 9 / ( 5 2 6 ) ( 5 2 6 ) 10 x x x x x x x x x x + − − − = − + = + + − = 2 1 2 2 7 2.3 2 2 4 1 1 10 / 6.(0,7) 7;11/ 1 1 ;12 / 3. 12 100 3 2 1 3 3 x x x x x x x x x x t t + + − − = + ≤ ≤ + = ÷ ÷ ÷ − − 2 2 2 2 2 1 2 2 2 3 6 3 5 2 13/ 9 9 10; 14 / 2 9.2 2 0; 15 / 2 15.2 2 ; 16 / 9 3 3 9 sin x cos x x x x x x x x x x x x + + + + − − + − + + = − + = + < − > − 2 1 /2 3 1 17 / 25 10 2 ; 18 / 4 2.6 3.9 ; 19 / 4.3 9.2 5.6 ; 20 /125 50 2 x x x x x x x x x x x x + + + = − = − = + = 3/ Sử dụng tính đơn điệu của hàm số: /2 1 1 1/ 4 2 4 1/ 2 1 3 ; 2 / 2 3 6 1; 3/ (2,5) (0,4) 2,9; 4 / 3 2 13; 5 / 2 6 x x x x x x x x x x x + + + + = + + > − + = + > = − 2 2 1 2 3 2 6 10 2 6 / 2 2 ( 1) ; 7 / 2 8 14; 8 / 3 6 6; 9 / 3 5 6 2 x x x x x x x x x x x x x x − − − − + − = − = − + − = − + − + = + 2 3 2 1 1 2 1 10 / 3 (3 10).3 3 0; 11/ 3.25 (3 7).5 2 0; 12 / (3 2 ) 2 2 0 x x x x x x x x x x x x − − − − + + − + − = + − + − = − − + − = 2 2 1 2 2 1 1 2 13/ 3 3 2 2 6 2 6; 3 3 2 14 / 2 3 5 2 3 5 ; 15 / 0 4 2 x x x x x x x x x x x x x x x − − − − − + + + − + − − = − + + − + + = + + ≥ − . 4/ Một số dạng khác: 2 2 2 2 2 3 2 6 5 2 3 7 3 2 6 5 1 2 1 1/ 4 4 4 1 (4 1)(4 1) 0; 2 / ( 2 1) 1 x x x x x x x x x x x x x x − + + + + + − + + + − + + = + ⇔ − − = − + ≥ 2 1 1 1 2 1 1 1 3 / 5.3 7.3 1 6.3 9 0 5.3 7.3 3 1 0; x x x x x x x− − + − − + − + − + = ⇔ − + − = 2 1 2 2 2 5/ 4 .3 4.3 1 0 4.3 4.3 1 (2.3 1) 0(*) x x x x x x + − + ≤ ⇒ − + = − ≤ ⇒ BPT vô nghiệm vì x = 0 KTM (*). 2 2 2 3 2 3 4 1 ( 1) 2 1 2 1 2 1 1 6 / 4 2 2 1; 7 / .2 2 .2 2 ; 8 / .3 (3 2 ) 2(2 3 ) x x x x x x x x x x x x x x x x x − + − + + − + + − − − + = + + = + + − = − 2 3 2 2 2 1 2 9 / .3 3 .(12 7 ) 8 19 12; 10 / 4 8 2 4 ( ).2 .2 . 2 x x x x x x x x x x x x x x x + + − = − + − + + − > + − + − 2 2 2 2 2 2 1 2 1 11/ 2 5 3 2 2 .3 . 2 5 3 4 .3 ; 12 / ( 1/ 2) ( 1/ 2) x x x x x x x x x x x x x x + + − − − + > − − + + ≤ + 2 2 2 2 10 10 2 11 20 13/ ( 4 ) (4 ) ( 10; 1;4); 14 / ( 2) ( 2) ( 1;2;3;4;5) x x x x x x x x x x x x − − + − − = − = ± − − = − = 2 3 1 2 2 2 3 2 2 2 1 5 15 /1/ (3 1) 1/ (1 3 ); 16 / ( 1) 1 ; 17 / ( 1) ( 1) + + + − + + − ≥ − − > − − + > − + x x x x x x x x x x x x x x II. Phương trình, bất phương trình lôgarít: 1/ Đưa về 1 cơ số: 5 25 0,2 2 3 4 20 1/ 3; 2 / 0,5 (5 4) 1 2 0,18; 3/ log x log x log log x log x log log x log x log x log x + = − + + = + + + = 2 5 1/5 5 1/25 4 / ( 6) 0,5 (2 3) 2 25; 5 / ( 1) 5 ( 2) 2 ( 2) log x log x log log x log log x log x + − − = − + + = + − − 3 2 3 3 2 2 3 3 1 1 1 4 4 4 3 1 3 6 / ( ). 1; ; 2 8 3 3 7 / ( 2) 3 (4 ) ( 6) 2 x log x log log log x x x log x log x log x − = + = ÷ ÷ + − = − + + 2 2 0,5 0,25 2 0,5 2 8 / ( 3) 5 2 ( 1) ( 1)( 2) 9 / (1 / 2) 2 / 4 0( 1) log x log log x log x log x log x + + = − − + − + − = − [ ] 5 5 5 5 25 0,2 2 12 / ( 6) ( 2); 13/ 3; 14 / ( 2 3) ( 3) / ( 1) 0 log x log x log x log x log x log log x x log x x = + − + + = + − + + − = 2 2 2 2 3 6 15 / 0,5. (5 4) 1 2 0,18; 16 / ( 1). ( 1) ( 1) log x log x log log x x log x x log x x − + + = + − − + − = − − 2 1/5 5 2 2 17 / ( 6 8) 2 ( 4) 0; 18 / ( 3) 1 ( 1) − + + − < + ≥ + − log x x log x log x log x 2/3 8 1/8 0,5 3 3 3 2 19 / 2 ( 2) ( 3) 2 / 3;20 / 1 ( (1 3) 1 0 3 ) log log x log x log x log x log x l og x − + − > + > ⇔ − > ⇒ < < 2 3 5 2 3 5 5 2 2 (2 ) 20 / . . ; 21/ 3 4. 5 1; 22 / ( 2). 2 2 0 x x log x log x log x log x log x log x log x log log x log − + + = + > + − ≥ ( 3) 0,25 2 2 3 3 2 23 / 6 2 (4 ) / ( 3) 1( 3); . 2 . 3 0(0 6 / 6; 1) x log log x log x x log x log x log x log x x x + + − + = = + ≥ < < ≥ 2/ Đặt ẩn phụ: 0,04 0,2 16 2 1/ (4 lg ) 2/ (2 lg ) 1; 2 / 1 3 1; 3 / 3 16 4 2 x x x log x log x log log x log x − + + = + + + = − = 2 3 2 1 2 2 1/ 2 4 / 16 64 3; 5/ l g(l g ) l g(lg 2) 0; 6 / (4 4). (4 1) 1/ 8 x x x x log log o o x o ox log log log + + = + − = + + = 2 2 (3 2 ) (3 ) 4 2 2 4 7 / (2 9 9) (4 12 9) 4 0; 8 / ( ) ( ) 2( 4 1) x x t log x x log x x log log x log log x x t − − − + + − + − = + = = ⇒ = 2 4 4 3 2 2 2 2 2 25 9 / (2 / ). 1( ( 1)( 2 2 1) 0); 10 / (125 ). 1(5&1/ 625) x x log x log x log x t t t t t log x log x + = ⇔ − + + + + = = 3 3 2 3 2 2 5 11/ 3 3 1/ 2; 12 / (4 1) (2 6) ; 13 / (5 ). 2 + + = + + + = − + = − x x x x x log log x log log x log log x log x log x 2 2 2 2 2 8 2 1 1 14 / 4 2 1 2 (2 1) 2 1 2 2 1 4 4 1: 3 / 2 2 1 x x log log log x x x t t t log x x + + − = + ÷ ÷ + − − ⇔ + = − − = ⇒ = ÷ + 2 2 1 18 / 10 6 0; 19 / l g(6.5 25.20 ) l g 25; 20 / 2(l g 2 1) l g(5 1) l g(5 5) x x x x log x log x o x o o o o − + + = + = + − + + = + 1/3 2 2 /16 2 21/ 5 / 2 3; 22 / 2. 2. 4 1; 23 / 2. 2 1/ ( 6) x x x x x log x log log log log x log log log x + ≥ > > − 2 3 3 3 2 4 1/2 2 16 24 / 4 9 2 3; 25 / 4 2(4 ) log x log x log x log x log x log x − + ≥ − + < − 1 2 1/2 2 2 2 2 1/2 4 26 / (2 1). (2 2) 2; 27 / 3 5( 3) x x log log log x log x log x + − − > − + − > − 2 2 2 2 2 2 3 5 2 6 4 6 28 / (2 ) (2 );29 / 2 2 (1/ 5); 30 / 4 2.3 ( 6 ) x x x log x log log x log log x log x log x log x log x x ≤ − ≥ − = = 3/ Phương pháp mũ hóa, lôgarít hóa: 5 11 1/ 4 4 2 3 5 5 2 3( 1) (lg 5)/3 5 lg 6 lg lg 3 1 1 2 ; 2 / 10 ; 3 / .5 11 ; 4 / 2 / ( 1 1) ( 1 1) x log log log x log x x x x x x x x x x x − − − + + + + − − = = = = + − − + + 2 2 1/3 4 9 2 2 (3 ) 5 / log ( 5) 0; 6 / (3 9) 1; 7 / ( 5 6) 1; 8 / (3 ) 1 x x x x x log x log log log x x log x − − > − < − + < − > 1/2 3 2 1/3 1/2 3 6 2 3 1 2 2 14 / 0; 15 / ; 1 2 2 1 3 2 16 / 0;17 / 1 2 2 x x x x x log log log log log log x x x x x log log log x x + + + − ≥ ≤ − − + − + > > + + [ ] 2 3 4 2 2 3 3 2 3 3/2 3 18 / ( ) 0; 19 / ( 2 3 ( 2)) t t log log log x log log x log log x t x t log log = = = ⇒ = = ⇒ = 2 3 3 2 3 3 2 3 3 3 2 3 3 20 / ( / ) 2 log log x log log x log log x log log x log log x log x log log x + = ⇔ = = ⇒ = 2 3 3 2 3 log log 2 3 2 3 3 2 3 2 21/ 3 2 (2 3) 1 x x t log log log log l og log x log log x log log t x ≥ ⇔ − ≥ − ⇔ ≥ ⇒ < ≤ 2log 3 2 3 3 2 3 log ÷ 2 2 3 4 4 3 2 3 2 3 4 2 3 3 22 / ( 4) ( ) (2 )( 1) log log log x log log log x x log log x log log x log t log t t = > ⇔ = ⇔ = > 3 3 3 3 1 4 2 48 1 48log log log t log x ∆ = + = ⇒ = + ⇒ = 1 48 3 3 4 log + ; 2 23/ ( 9 1) 1 x log x x − − − ≥ 2 2 4 0,5 1,5 2 0,25 2 24 / 2. 2 ; 25 / ( 3) 1 ( 4) ( / 3) 0 x log x log x x log x log x x log log x x − ≥ + + ≥ ⇔ − + ≥ 4/ Sử dụng tính đơn điệu của hàm số: 2 3 5 1/ l g( 6) 4 l g( 2) l g( 3) 4 4; 2 / ( 1) (2 1) 2 2 x o x x o x x o x x log x log x x + − − = + + ⇔ + − = ⇒ = + + + = ⇒ = 2 3 3 3 3 / ( 2) ( 1) 4( 1) ( 1) 16 0( ( 1) 4; 4 / ( 2) 80 / 81;2) x log x x log x log x x x + + + + + − = + = − + ⇒ = − 2 2 2 2 3 9 3 2 2 4 / (1 ) ( 1 3 2 2); 5 / .3 ( 9 12 3 ) t t log log x log t t t log x log x t t x x x log x t + = = ⇒ + = ⇒ = = − = ⇒ = − 5 6 3 3 2 ( 3) 5 2 6 / 3 (1 ) 2 ( 2 1 8 4 9 2); 7 / 2 ( 3) t t t t log x log x x log x x t x log x log x + + + = = ⇒ + + = ⇒ = = ⇔ + = ( ) 6 6 6 2 6 2 3 5 2;8 / 3 3 2 6 3 2 1 1/ 6 log x log x log x t t t t t t x log x log x x t x = ⇒ + = ⇒ = + = ⇔ + = ⇔ + = ⇒ = − ⇒ = 2 2 2 1 3 2 2 3 9 / 3 1;10 / 2 2 8 / (4 4 4)( , 1/ 2) log x x x x log x x VP VT x + − = − + = − + ≤ = 7 3 7 3 11/ ( 2)( ( 7 2) 3 7 2 1 (2) ( ) t t t log x l og x log x t t log f f t < + = ⇒ < + ⇔ < + ⇔ = < = 7 ( 7 /3) 2.(1/3) 2 49 0) t t t log x x + ⇒ > = ⇒ > > 2 2 2 2 2 1/3 3 2 3 2 3 12 / 4 ( 2 3) 2 (2 2 2) 0 2 ( 2 3) x x x x x log x x log x log x x − − − + − + − + + − + = ⇔ − + = 2 2 2 2 3 2 (2 2 2) 2 3 2 2 2 3 x log x x x x x − + − + ⇔ − + = − + ⇒ = − 2 2 2 2 3 2 2 3 13 / ( 5 5 1) ( 5 7) 2( 5 5 ( ) ( 1) ( 2) 2 log x x log x x t x x f t log t log t − + + + − + ≤ = − + ⇒ = + + + ≤ (1) 0 1 1 (5 5)/ 2 (5 5)/ 2 4f t x x = ⇒ ≤ ≤ ⇒ ≤ ≤ − ∧ + ≤ ≤ [ ] 2 1/2 1/2 2 2 16 / ( 1) (2 5) 6 0 ( 2) ( 1) 3 0 0 2 4 x log x x log x log x x log x x x + + + + ≥ ⇔ − + − ≥ ⇒ < ≤ ∧ ≥ 2 2 2 3 5 11 2 2 5 11 ( 4 11) ( 4 11) 17 / 0 2 5 3 ( 4 11 0; ( ) 2 3 ; log x x log x x x x t x x f t lo g t log t − − − − − ≥ − − = − − > = − 2 2 2 1/3 1/3 2 19 / ( 1) 2( 3) 8 0;20 / 2 8 (2 1) / ( 1) + + + + ≤ − = + − ⇔ x log x x log x x x log x x 2 2 2 2 2 2( 1) ( 1) (2 1) 2(2 1) ( 1) 2 1 0;4x log x log x x x x x − + − = + + + ⇔ − = + ⇒ = 5/ Một số Phương trình, bất phương trình khác: [ ] 2 2 2 1/3 1/3 2 ( 6) 1/1/ 2 3 1 1/ ( 1) (0;1/ 2) (1;3/ 2) (5; ) ; 2 / (2 3.2 ) 1( 1) log x log x x x log x x log x a − + − − + > + ∪ ∪ +∞ + > > 3/ ( 1) lg1,5(0 1 0 ; 1 1 ) x log x x VT VP x VT VP + = < < ⇒ < < > ⇒ > > 2 2 2 2 2 2 4 / ( 3 1) 2 0 0 ( 2)( 3) 1& 3 2 3 1 0 log x x log x t t t t t x + − − + ≤ ⇔ < − + − ≤ > ⇒ > > ⇒ > > 1 2 2 3 3 5 / (3.2 1) / 1( 1 (2 / 3) 0); 6 / ( 1) / (9 3 ) 3 1( log 9 3 0) x x log x x log x x log MS − − ≥ ≥ ∧ < < − − − ≤ < − < [ ] 5 5 3 3 4 4 7 / ( / 3) (2 ) / ((0; 5 / 5) (1;3)); 8 /1/ ( 1) / ( 2) 1/ ( 3) x log x log x log x log x log x log x x log x + < − ∪ + + < + III. Hệ phương trình, bất phương trình mũ và lôgarít: 2 3 2 1 3 1 2 3 2 3 2 77 2 5 4 2 2 12 1/ ;2 / ;3/ ; 5 3 2 7 4 2 (2 2) 2 2 3.2 4 / ; 3 1 1 x y x x y x y x x x x y y x y y x y y x xy x + + − + − = = − + = + = − = + = + + = + + = + 1 2 1 4 5( ) 26 4 3.4 2(1) 6 / ; 7 / 3 2 3(2) 64 x y y x y log y l og x x y log xy + − − + = + ≤ + ≥ − = 8 8 12 2 2 2 x+y 3 3 3 3 ( ) 4 4 3 8 / ; 9 / ; 12 (1 2 2) 1 4 10 / ; 11/ 1 2 3 x y log y lo g x x y xy xlog log y y log x x y xlog log x y log y y x log log xy x y log x log y x y + + + = + = + = + = + = + = − = − = 2 2 2 2 2 2 1 2 2 2 2 3 2 ( )( 1) 12 / ; 1 0(1) .2 3 .2 2 13/ ;14 / 2 .2 3 .8 1 / 3 3 5 9 0(2) x y x y x y x y x y e e log y log x xy x y log x log x x y x y x x x − + + + + − = − + + = − < + = + = − + + > 1/4 4 2 2 2 3 9 3 1 ( ) (1/ ) 1 1 2 1 15 / ; 16 / 25 3 (9 ) 3 ( 1)lg 2 lg(2 1) lg(7.2 12) 17 / ( 2) 2 + − − = − + − = + = − = − + + < + + > x x x log y x log y x y x y log x log y x log x . Chuyên Đề 2: ph ng trình logarit ơ I. Phương trình, bất phương trình mũ : 1/ Đưa về cùng một cơ số hoặc hai