# Exponent & Logarithm

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1. Exponent Formula 2. Consider the following the problem What’s the product of 24 and 25 ? Solutions: am = a.a.a…..a m times 24 = 2.2.2.2 =16 4 times 25 =2.2.2.2.2…
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• 1. Exponent Formula
• 2. Consider the following the problem What’s the product of 24 and 25 ? Solutions: am = a.a.a…..a m times 24 = 2.2.2.2 =16 4 times 25 =2.2.2.2.2 =32 So that the answer of the problem above is: 24 .25 = 16.32 = 512 We know : 512 = 29 =16.32 =24 .25 = 2 4+5
• 3. If we replace the base a by a, the exponents 4 and 5 by positive integers m and n : We will get: am = a.a.a….a m times an = a.a.a….a n times am .an = (a.a….a).(a.a….a) m times n times = a.a.a……..a (m+n) times FORMULA I with condition a € R, a ≠ 0 a is called the base number and m,n is called exponent .m n m n a a a + =
• 4. if m>n, them use formula I, we shall obtain: am-n .an = am-n+n am-n .an = am Formula II m m n n a a a − = an
• 5. pay attention the following problem : (am )n = am .am …..am n times = am+m+m….m = am.n Conclusion Formula III . ( ) ( )m n n m m n a a a= =
• 6. If besides a we also use the base of b, then the form (ab)m . Can be defined : (ab)m = ab.ab……ab m times = (a.a.a.a).(b.b.b.b) m times m times Formula IV ( ) .m m m ab a b=
• 7. So that by the same way we will get Conclusion : b≠ 0 Formula V Because : is not defined m m m a a b b   = ÷   0 number
• 8. Example : Simplify! ( ) 3 5 2 5 2 4 2 3 54 11 8 19 6 5 11 1.5 .5 .5 2.( ) . . . 3.2 .(2 ) .(2 ) 4.( 3) . 3 3 3 3 5. 3 3 3 x x x x x x x − − − =
• 9. ( ) ( ) ( ) ( ) ( ) ( ){ } ( ){ } ( ) ( ){ } ( ) ( ) ( ) 8 9 7 2 85 2 53 4 32 3 5 3 42 2 3 4 42 3 32 3 . 3 6) 3 . 3 7) 8) 2 . 2 3 . 3 9) 3 10) . . 11) . a a a a p y y y y y x x y y x y x y − − − − −      ÷  ÷     −
• 10. ( ) 3 5 2 10 5 2 4 12 2 3 2 2 3 3 6 6 54 9 11 8 19 19 11 8 6 5 11 1.5 .5 .5 5 2.( ) . . . 3.2 .(2 ) .(2 ) 2 .2 .2 2 4.( 3) . 3 3 3 3 3 5. 3 3 3 3 3 x x x x x x x x x x x x − = − = − = = − − = − = = =
• 11. ( ) ( ) ( ) ( ) ( ) 8 9 17 17 8 8 7 2 9 9 85 5.8 40 3 . 3 3 . 6) 3 . 3 .3 . 3 7) a a a a aa a p p p − = = − −− − = = ( ){ } ( ){ } ( ) ( ) ( ) ( ) 2 5 2 53 4 3 3 4 4 6 6 20 20 26 26 8) 2 . 2 2 . . 2 . 2 . . 2 . 2 . y y y y y y y = = =
• 12. ( ) ( ){ } ( ) ( ) ( ) 32 3 32 2 3 3 5 5 5 2 2 9 9 5 5 11 11 5 5 6 6 3 . 3 3 . . 3 . 9) 3 .3 3 . . 3 . 3 . 3 . 3 . 3 . y y y y yy y y y y y y − − − = − = − = = − ( ) ( ) 3 42 2 6 8 14 3 4 9 16 25 42 3 8 12 5 6 3 3 62 . 10) . . . . 11) . .. x x x x x y y y y y x y x y x y x yx y     = = ÷  ÷     = = − −−
• 13. In this section we shall see that formula I-V hold for integers exponents, either positive, negative and zero If we substitute m = n use formula II we shall obtain Formula VI a € R, a ≠ 0 m n n n n n a a a a a − = = 0 1 a=
• 14. If we subsituty m=0 use formula II we will get Formula VII a € R, a ≠ 0 0 1m n n n a a a a a = = 0 1n n a a − = 1n n a a − = 1 n n a a− =
• 15. ( ) ( ) ( ) 5 2 7 5 2 2 3 4 25 3 3 42 2 1)3 .3 4 2) 4 3) . 4) 5) 6) . x x x x x x x y x y − − − − − − − − − SIMPLIFY!!
• 16. 3 3 22 2 3 7) . 8) x y ab b c − − − − −    ÷   ( ) ( ) ( ) ( ) 4 3 1 2 2 1 . 1 2 9) 3 . 3 x x x x − − − − − + + 3 13 4 4 2 3 5 10) : a b a b c b c − − −    −  ÷  ÷    
• 17. ( ) ( ) ( ) ( ) 5 2 3 7 12 5 2 2 0 3 1 4 25 10 13 3 3 13 3 4 12 2 2 2 1)3 .3 3 4 2) 4 4 3) . 1 4) 1 5) 1 6) . x x x x x x x x x x x x x y x y x y x y − − − − − − − − − − = = = = = = = = − − = − = −
• 18. ( ) ( ) 33 3 3 22 2 4 2 3 4 6 2 8 6 1 7) . 1 8) x y xy xy ab a b b c b c a b c −− − − − − − − = =   = = ÷   ( ) ( ) ( ) ( ) ( ) ( ){ } ( ) ( ) ( ) ( ) ( ) 344 3 1 2 3 1 3 3 2 1 2 12 1 . 1 2 9) 3 . 3 3 2 1 3 3 2 1 x xx x x x x x x x x −− − − − − − − − −− − = + + + − − = + + = − −
• 19. 3 13 4 4 9 12 4 1 2 3 5 6 3 5 9 12 3 5 6 4 1 9 15 6 4 1 5 5 14 11 5 14 11 10) : : : . . . . . . . 1 a b a b a b a b c b c c b c a b b c c a b a b c a b c a b c a b c − − − − − − − − − − − − − − − − − − − − − −    − − = ÷  ÷     − = = − = − = −
• 20. By the Formula I we obtain am .an = am+n If we substituting n=m, then we will get am .am = a2m so that a2m = a 2m =1 m =1/2 form a2m =a ,can be changed (am ) 2 = a am = √a a1/2 = √a In general can written 1 nn a a=
• 21. In the fractional exponents: If is changed ,then we will get : n,m is positive integers Formula VIII 1 n m n ( ) 1 mm n n m m nn m n mn aa a a a a   =  ÷ = =  
• 22. By the formula VIII: ,then if ,we obtain Formula IX Furthermore the formula I - IX,also hold for fractional exponents 1 n=n n p a a q − = 1 1 orp q p p q q p q a a a a − − = = 1 1 orm n m m n n m n a a aa − − = =
• 23. Example 1. Simplify and write down in positive exponents! ( ) ( ) ( ) 131 3 62 2 2 1 3 6 31 4 4 2 3 . 2 .2 . 2 . . . 3 . 3 . 3 . a b x x x x x c x a d b − − −    ÷   5 35 6 1 5 2 3 34 3 .2 . . . x x e x x x f x −− −    ÷  ÷  
• 24. 2. Simplify and write down in the roots form ! ( ) ( ) 1 2 34 . 1 3 . 3 1c x x− − ( ) 1 3 3 1 1 6 2 5 45 1 52 . . 1 . 2. 4 . x x a x x x b x x − −
• 25. ( ) ( ) 2 3 4 33 3 4 3 42 3 . 27 1 . 2 .2 16 . 81 1 . 125 81 a b c d − − − −    ÷     + ÷   ( ) 1 2 2 2 3 2 1 5 34 . 3 4 .16 32 125 e f − + + − 3. Evaluate!
• 26. 4. Find the length of diagonal of rectangle if it has dimension length is cm and width is cm! 8 2
• 27. ( ) ( ) ( ) ( ) ( ) 13 31 1 1 1 3 62 2 2 2 2 2 1 2 2 1 2 1 41 3 6 3 3 3 31 4 4 31 1 0 4 4 2 22 3 3 1 1. . 2 .2 . 2 2 2 2 . . . 3 . 3 . 3 3 1 3 . a b x x x x x x x c x x x a a d b b − − −+ − + − − − + − = = = = = = = =     = ÷  ÷    
• 28. 31 4 5 35 5 5 5 15 35 6 11 11 1 5 55 22 3 3 32 3 2 3 3 334 4 2 2 3 .2 6 . 6 . 6 6 . 1 . x x x x x e x x x xx x x x x f x x x x x − −−− −− − − − = = = =     ÷= = = = ÷  ÷  ÷    
• 29. ( ) 1 3 1 3 3 32 3 1 1 12 2 43 6 62 1 1 1 6 6 62 2 4 5 45 5 5 5 1 1 61 1 52 52 52 . . 2. . 1 1 1. . 2 2 2. 1 4 . 4 . . 4 . x x x x a x x x x xx x x x x x b x x x x x + + −− − − − = = = = = = = = ( ) ( ) ( ){ } ( ) ( ) ( ) 11 22 434 3 11 12 11 12 . 1 3 . 3 1 3 1 . 3 1 3 1 1 3 c x x x x x x − − = − − − = − − = −
• 30. ( ) ( ) ( ) ( ) 22 3 233 6 23 64 2 33 33 4 3 344 4 3 3 3 3 324 42 3 3 23 4 1 3. . 27 3 3 9 1 1 1 1 1 . 2 422 .2 2 16 2 2 3 27 . 81 3 3 2 8 1 1 1 1 1 . 125 5 5 27 27 81 3 3 25 25 a b c d −− − −− − − − − −−− − = = = = = = =    = = = = ÷ ÷           + = + = + = + = ÷  ÷  ÷      
• 31. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 11 1 2 2 2 12 22 2 3 2 13 2 1 4 5 34 5 35 34 3 2 1 . 3 4 9 16 25 5 5 5 .16 32 125 2 2 5 2 2 5 8 4 5 7 e f − −− − − + = + = = = = + − = + − = + − = + − =
• 32. ( ) ( ) 2 2 2 2 2 4. 8 2 . 8 2 10 10 x x x x = + = + = =
• 33. f(x) p I. a =a f(x)=p⇒ Example : 1. Find the solution set : 3 1 1 )2 8 2 1 1 ) 27 3 243 x x a b − =   =  ÷  
• 34. 3 1 )2 8 2 x a = 1 1 1 ) 27 3 243 x b −   =  ÷  3 3 1 2 1 3 2 2 2 .2 2 2 1 3 2 1 6 x x x x − ⇔ = ⇔ = ⇔ = ⇔ = 3 1 52 3 1 52 3 .3 3 3 3 3 1 5 2 3 6 2 3 12 4 x x x x x x − − ⇔ = ⇔ = ⇔ − = ⇔ = ⇔ = ⇔ =
• 35. f(x) ( ) II. a =a f(x)=g(x)g x ⇒ Example : 1. Find the solution set : ( ) 3 1 4 3 5( 1) 2 3 2 3 4 8 16 1 ) 1 162 2 )27 3 .81 ) 3 27 xx x x x x a b c − + − + −   =  ÷   = =
• 36. 3 1 4 16 1 ) 1 162 2 xx a −   =  ÷   3 4 4 4 4 1 1 2 4 4 4 1 3 4 2 4 4 3 2 2 2 .2 2 2 .2 1 2 3 4 4 3 2 5 7 2 5 14 x x x x x x x x x x − − − − − − − ⇔ = ⇔ = ⇔ = − − ⇔ − = − − ⇔ = ⇔ =
• 37. 3 5( 1) )27 3 .81x x b + − = ( ) 2 3 2 3 4 8 ) 3 27x x c + − = ( ) ( )2 2 3 3 4 8 3 2 3 3 4 6 12 24 3 2 8 12 36 72 84 28 3 x x x x x x x x + − ⇔ = + − ⇔ = ⇔ + = − ⇔ = ⇔ = 3( 3) 5( 1) 4 3 3 .3 3 9 5 5 4 3 9 5 1 10 2 5 x x x x x x x x + − ⇔ = ⇔ + = − + ⇔ + = − ⇔ = ⇔ =
• 38. III. Exponent equation is changed to quadratic equation Example : 1. Determine the solution set : ( ) 1 2 3 )3.9 10.3 3 0 )2 2 36 )5 6 5 5 0 x x x x x x a b c + + − + = + = − + =
• 39. )3.9 10.3 3 0x x a − + = ( ) ( ) ( ) 2 2 3. 3 10.3 3 0 3 , 0 3 10 3 0 3 1 3 0 x x x a a a a a a ⇔ − + = ⇔ = > ⇔ − + = ⇔ − − = 1 3 1 3 3 1 x a x ⇔ = ⇔ = ⇔ = − 3 3 3 1 x a x = = = V
• 40. 1 2 3 ) 2 2 36x x b + + + = ( ) ( ) 2 3 2 2 2 , 0 2 .2 2 .2 36 0 2 8 36 0 4 18 0 4 9 2 0 x x x p p p p p p p p ⇔ = > ⇔ + − = ⇔ + − = ⇔ + − = ⇔ + − = : 2 9 4 p⇔ = − 2 2 1 x x ⇔ = ⇔ = V 2p =
• 41. ( ))5 6 5 5 0 x x c − + = ( ) ( ) 2 5 5 2 x x = = ( ) ( ) 0 5 5 0 x x = = V V ( ) ( ) ( ) ( ) ( ) 2 2 5 6 5 5 0 5 , 0 6 5 0 5 1 0 5 1 x x x p p p p p p p p ⇔ − + = ⇔ = > ⇔ − + = ⇔ − − = ⇔ = = ⇔
• 42. (1) Rational Number Rational Number are numbers which can be put in a/b form with a and b are integers numbers and b ≠ 0. Example: 1. 3 coz 3 can be stated in etc 2. 0.444…= 0.4 can be stated in 3. 1.12 can be stated in Notes ! The lines above 12(no.3)shows that 12 is repeated unlimitedly Generally ! Rational number are repeated decimals 6 9 , 2 3 2 5111 97
• 43. (2) Irrational Numbers Irrational number are number which can’t be stated in Generally : Irrational numbers are unrepeated decimal Example! 1. =1.4142135…. 2. = 5.1461524…. 3. Log 2 = 0.3010…. 4. e =2.71828182….. coz can’t be put in 2 a b 3 3 a b
• 44. 1.Please check whether the following form are rational or irrational. 3 3 a.-2 5 1 .log 10 . log 2 . 12 b c d − . 0.1 2 . 9 .1.2 .8.34 e f g h
• 45. ( ) ( ) ( ) ( ) 3 3 a.-2 5 1 .log 10 . log 2 . 12 I b R c I d I− ( ) ( ) ( ) ( ) . 0.1 2 . 9 .1.2 .8.34 e I f I g R h R
• 46. Root number are the rational number which the result are irrational EXAMPLE : 3 3 6 1) 3,2 12, 8 are root forms 2) 4, 27, 64, , are not root form because the result are rationalπ−
• 47. 3 5 )2 7 ) log81 ) 0,05 9 ) 25 ) 32 a b c d e − 1. Please check whether the following forms are root forms or not 5 )1, 2 ) 81 1 ) 3 ) )ln 2 f g h i e j Note : Ln = Log number with base e Ln x = e log x
• 48. 2. :Simplify the following root form 4 5 5 44 ) 32 )4 250 ) 243 ) 64 1 ) 288 2 ) 405 a b c d e f x y −
• 49. ( ) ( ) 511 54 4444 ) 32 16.2 4 2 )4 250 4 25.10 20 10 ) 243 243 3 3 3 3 a b c = = = = = = = = ( ) 5 55 5 5 4 4 4 54 4 ) 64 2 .2 2 2 1 1 1 ) 288 144.2 .12 2 6 2 2 2 2 ) 405 81.5 . . 3 5 d e f x y x x y xy x − = − = − = = = = =
• 50. ( ) ( ) 2 2 Notes ! 1) 2 2) 2 a b a b ab a b a b ab + = + + − = + − ( ) 2a b a b ab+ = + + ( ) 2a b a b ab a b− = + − ⇒ ≥
• 51. ( ) ( )3. inSimplify a b or a b form+ − ) 8 2 12 ) 15 2 90 ) 9 80 ) 14 6 5 ) 15 10 2 5 2 ) 6 3 a b c d e f − + − + − + ANSWER
• 52. ( ) ( ) ( ) ) 8 2 12 6 2 2 6.2 6 2 ) 15 2 90 8 7 2 56 8 7 ) 9 80 9 2 20 4 5 2 5 ) 14 6 5 14 2.3 5 14 2 45 9 5 2 9.5 9 5 ) 15 10 2 15 2 50 10 5 5 2 5 1 5 1 1 1 1 1 1 1 ) 4. 2 2 . 6 3 6 6 6 6 2 3 2 3 2 3 a b c d e f − = + − = − + = + + = + − = − = − = − + = + = + = + + = + − = − = −     + = + = + = + + = + ÷  ÷    
• 53. REMEMBER !! 1) . 2) . 3) 4) a a a a b ab a b a b a b a b = = + = + − = −
• 54. 1. Evaluate the value of the following root forms ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 )5 2 2 2 2 )20 3 80 45 ) 150 3 54 5 96) ) 2 8 ) 6 4 6 4 1 1 ) 3. 27.10 48 2 3 ) 4 8 3 4 8 3 ) 1 5 10 1 5 10 a b c d e f g h + − − − − + + − + + + − − + + − − − ANSWER
• 55. ( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 )5 2 2 2 2 5 3 1 2 7 2 )20 3 80 45 2 5 3.4 5 3 5 2 12 3 5 7 5 ) 150 3 54 5 96) 5 6 3.3 6 5.4 6 5 9 20 6 16 6 ) 2 8 2 8 2 16 18 ) 6 4 6 4 6 4 10 1 1 1 1 1 ) 3. 27.10 48 3. .3 3.10.4 3 .40 3 20 3 2 3 2 3 2 a b c d e f + − = + − = − − = − + = − + = − − + = − + = − + = + = + + = − + = − = − = = =
• 56. ( )( ) ( ){ } ( ){ } ( ) ( ) 2 2 ) 4 8 3 4 8 3 4 8 3 4 8 3 4 8 3 16 8 3 2 8. 3 5 2 24 5 4 6 g + + − − = + + − − = − + = − + + = − = − ( ) ( ) ( ){ } ( ){ } ( ) ( ) 2 22 2 2 2 ) 1 5 10 1 5 10 1 5 10 1 5 10 1 5 10 1 5 10 2 5. 10 16 2 50 16 10 2 h + + − − − = + + − − − = + + = + + + = + = +
• 57. 1. Find the value of the following problem !! ) 10 10 10 10... ) 72 72 72 ... ) 56 56 56 56 ... a b c + + + − − − − 2+ 5 2 5 if p q = = − 2. Find 2p+2q, 4pq and p2 +q2 ANSWER
• 58. 1 ) 10 10 10 10...a 10 10 10 10...x = 0 10x x= ∨ = 10x∴ = 2 10 10 10 10...x⇔ = 2 10x x⇔ = ( ) 2 10 0 10 0 x x x x ⇔ − = ⇔ − =
• 59. 1 ) 72 72 72 ...b + + + 72 72 72 ...x = + + + 9 8x x⇔ = ∨ = − 9x∴ = ( ) ( ) 2 2 2 72 72 72 ... 72 72 0 9 8 0 x x x x x x x ⇔ = + + + ⇔ = + ⇔ − − = ⇔ − + =
• 60. 1 ) 56 56 56 56 ...c − − − − 56 56 56 56 ...x = − − − − 8 7x x⇔ = − ∨ = 7x∴ = 2 56 56 56 56 ...x⇔ = − − − − 2 56x x⇔ = − ( ) ( ) 2 56 0 8 7 0 x x x x ⇔ + − = ⇔ + − =
• 61. ( ) ( )2 ) 2 2 2 2+ 5 2 2 5 4 2 5 4 2 5 8 a p q+ = + − = + + − = ( )( ) ( ) 2 ) 4 4 2+ 5 2 5 4 4 5 4 b pq = − = − = − ( ) ( ) ( ) ( ) 2 2 2 2 2 ) p +q = 2 5 2 5 4 4 5 5 4 4 5 5 4 4 5 5 4 4 5 5 18 c + + − = + + + − + = + + + − + =
• 62. A fractional of root on its denominator such as : 1 , , c c a a b a b+ − Can be simplified by rationalizing 1 1 1 . a a a a a a = = ( ). c a bc c a b a ba b a b a b −− = = −+ + − ( ). c a bc c a b a ba b a b a b ++ = = +− − +
• 63. 2 2 3 2 3 ) . 33 3 3 a = = 5 5 12 5.2 3 15 ) . 12 12 612 12 b = = = 4 4 3 2 4 3 4 2 ) . 4 3 4 2 3 23 2 3 2 3 2 c − − + − − = = = − − −− − +
• 64. ( ) 2 2 2 3 2 3 8 2 2 24 4 3 ) . 8 2 8 2 8 2 8 2 4 6 4 3 6 3 4 d − − = = + + − − − = = − ( ) ( ) ( ) 2 2 2 6 26 2 6 2 6 2 ) . 6 2 6 2 6 2 6 2 6 2 2 12 8 4 3 2 3 6 12 4 e −− − − = = + + − − + − − = = = − −
• 65. Logarithm Formula Excercise
• 66. a c 1. log b c a b= ⇔ = a 2. log a 1= a 3. log 1 0= a a a 4. log x . y log x log y= + a a a 5. log log x log y x y = − a n a 6. log x n . log x=
• 67. x log 9. log y log a a x y = a y a 10. log b log b x y x = x a y 8. log a y x = x 1 7. log y logy x = BACK 11. log . log . log loga x y a x y b b= log 12. a x a x=
• 68. 2 2 2 3 log 2 log 3 3 3 ½ 2 2 ) log 16 1 ) log 25 ) log 0.001 ) ( 3) 1 ) ( ) 4 ) log 2 log 4 5 ) log 5 - log 4 a b c d e f g + 3 4 3 2 7 9 3 2 25 4 2 8 4 ) log 3 ) log 7 . log 27 . log 2 log 8 ) log 4 1 ) log3 log 4 7 7 ) log 21 log7 log . 24 3 h i j k l −   − −  ÷ ÷   ANSWER 1. EVALUATE !!
• 69. 2 2 4 ) log 16 log 2 = 4 a = 5 5 2 5 -2 1 1 ) log = log 25 5 log 5 2 b = = − -3 ) log 0.001 = log 10 -3 c = 1 3 log 2 3 log 22 1 3 log 2 2 1 2 ) ( 3) (3 ) = (3 ) 2 2 d = = = 2 2 log 3 log 3 -2 -2 1 ) ( ) = (2 ) 4 = 3 1 9 e =
• 70. 3 3 ½  3 3 3 3 )      log 2   log 4   =  log 2 +  log 2 =   log 2.2   log 4 f + = 2 2 2 2 5 4 5 5 )     log 5 -   log   =    log  4    log 4 =  2 g = 13 3 4 4 4 3 )     log       log  3 4 =  -1 h −   =  ÷  
• 71. 3 2 7 3 7 2 3 )      log 7 .  log 27 .   log 2 =   log 7 .   log 2 .   log 27 =   log 27 = 3 i 2 1 2 9 3 3 3 3 2 3 3 1 2  log 8  log  2 )     =   log 4 log  2 3   log 2 2    2   log 2 3 1 =   .  2 4 3   8 j = =
• 72. 2       2 25 5 5 2 5 5 5 5 1 1 1 ) log5  =  log 4 log 2 log 2 1 1   log 2 log 2 1   log 2 log 2  1 k x − − = − = =
• 73. 2 3 4 2 8 1 2 2 2 2 13 2 3 2 2 2 2 3 3 2 1 3 3 1 2 2 2 22 3 3 2 7 7 )      log21    log7 log . 24 3 7 7   log  21    log7   log . 2 .3 3 1 1 7     log  21 -  log7 -    log   2 3 2 .3 7   log  21   log7 -  log   2 .3 l   − −  ÷ ÷      ÷= − −  ÷  ÷   =    ÷= −  ÷  ÷   { 1 222 2 1 2 1 2 2 2 1 2 2 2 7 2 3 7    log  21   log7 log 2.3 7    log  21 log7 .  2.3 7 3    log 7 3    log  14 } x x    ÷= − +  ÷  ÷      ÷= −  ÷  ÷   = = NEXT
• 74. 2. SIMPLIFY !! ( ) ( ) 2 2 2 23 1 1 ) log x + log   - log  ) log log ) log log logx x x a x x b x y x y c x x x − − − + − ANSWER
• 75. 2 2 2 2 1 1 1 ) log x + log   - log   = log x . log   . log x 1 log x .   . x log  x 2 log x a x x x x   =  ÷   = = ( ) ( ) ( ) ( ) ( )2 2 ) log log log log log log . log log . { }b x y x y x y x y x y x x x y y y x y − − − = + − − − = + − =
• 76. 1 1 3 2 23 x log  .  log ) log log log 2  logx 1 1 . 3 2 2 1 12 x x x x x x x c x x x+ − = = = NEXT
• 77. 3. Given that : log 2 = 0.301 log 3 = 0.477 log 4 = 0.845 EVALUATE !! )log5 7 )log 2 6 )log 7 a c e 12 )log6 3 )log 2 ) log 3 b d f ANSWER
• 78. 10 )log5 log 2 log10 log 2 1 0.301 0.699 a = = − = − = )log6 log 2 .  log3 0.301 0.477 0.778 b = = + = 7 )log log7 log 2 2 0.845 0.301 0.544 c = − = − = 3 1 1 )log log3 log 2 2 2 2 1 1  . 0,477    . 0,301 2 2 0.2385 0,1505 0,088 d = − = − = − = NEXT
• 79. 4 9 3 9 4. Given that : )  log 6 = A      Express  log 8 with A b)  log 5 = B      Express  log 375 with B a 3 2 6 )  log  2 = M      log 7 = N     Express  log 98  c ANSWER
• 80. ( ) 4 9 4 4 3 4 2 4 4 log8 )  log 8 =  log9 log 2 log3 3 .  log 2 2 .  log3 13 .  2 12 .  2 3 22  .  2 1 2 3 4 2 a A A A = = = − = − = −  : 4log 6 = A 4log 3 . 2 = A 4log 3 + 4log 2 = A 14log 3 +   = A 2 14log 3 = A -  2 Note
• 81. 3 9 3 3 3 3 3 3 log  375 )  log  375 =  log  9 log  125 . 3 2 log  5  +  log 3 2 3  log5 + 1 2 3 1 2 b B = = = + =
• 82. 3 6 3 3 3 3 2 3 3 3 3 3 2 log98 )  log 98 =  log6 log 49.2 log3.2 log7 log 2 log3 log 2 2  log7 1 2  log 2  log7 1 2 1 c M M M M MN M M = + = + + = + + = + + = + NEXT
• 83. 5. Find the value of x that fulfill of the following equation! ( ) ( ) ( ) 3 3 3 3 3 3 4 4 4 4 4 )  log log 1 log 2 )   log 2 1 log 3 log7 )  log .  log x -  log log log16 = 2 a x x b x x c + + = − − − = ANSWER
• 84. ( )3 3 3 )  log log 1 log 2a x x+ + = 2 1x x= − ∨ = TM { }1HP = ( ) ( )3 3 3 )   log 2 1 log 3 log7b x x− − − = { }4HP = ( ) ( ) ( ) ( ) 3 3 log 1 log 2 1 2 2 1 0 x x x x x x ⇔ + = ⇔ + = ⇔ + − = ( ) ( ) ( ) ( ) 3 32 1 log log7 3 2 1 7 3 2 1 7 21 4 x x x x x x x − ⇔ = − − ⇔ = − ⇔ − = − ⇔ =
• 85. 4 4 4 4 4 )  log .  log x -  log log log16 = 2c { } { } 4 4 4 4 2 4 2 4 2 2 16 16 log 2  log x 2 log   log x log4 log x 16 x 4 2 65536 x x ⇔ = ⇔ = ⇔ = ⇔ = ⇔ = ⇔ = ( ) ( ){ } ( ) 4 4 4 4 4 4 4 4 4 4 4 4 4 4 log  log x -  log log log16  2 log x log 2 log log16  log x log 2 log2 log x log 2 1 2 ⇔ =    ⇔ =       ⇔ =       ⇔ =    
• 86. 1. Find the value of x that fulfill of the following equation !! ( ) ( ) ( ) ( ){ } { } ( ) 2 2 5 2 5 ) log 2 3 4 log 2 8 ) log 2 log 5 1 )log log 3 log 2 log log16 ) log 2 2x a x x b x x c x x d x − + = − − + = + + = + = ( ) ( ) ( ) 5 2 5 3 32 2 2 ) log 12 3. log 4 1 0 ) log log 2 0 ) log 1 log 1 10 0 x x e x f x x g x x + − + = − + = + − + − = ANSWER
• 87. ( ) ( )2 2 ) log 2 3 4 log 2 8a x x− + = − ( )5 2 5 ) log 2 log 5 1b x x− + = 5 10 2 x x⇔ = − ∨ = HP = 5 ,10 2 HP   = −    Ø ( ) ( ) ( ) ( ) 2 2 2 2 2 log 2 3 log16 log 2 8 log16 2 3 log 2 8 32 48 2 8 1 1 3 x x x x x x x ⇔ − + = − ⇔ − = − ⇔ − = − ⇔ = ( ) ( ) ( ) 5 2 5 5 2 2 2 log 2 log 5 log5 2 5 5 2 5 25 2 5 25 0 2 5 5 0 x x x x x x x x x x ⇔ − + = ⇔ = + ⇔ = + ⇔ − − = ⇔ + − =
• 88. ( ){ } { })log log 3 log 2 log log16c x x+ + = 1 9x x⇔ = ∨ = ( )) log 2 2x d x + = 2 1x x⇔ = ∨ = − { }2HP = { }1,9HP = ( ){ } { } ( ){ } ( ) ( ) ( ) 2 2 2 log log 3 log 2 log log16 2.log 3 log16 log 3 log16 6 9 16 10 9 0 1 9 0 x x x x x x x x x x x x x ⇔ + + = ⇔ + = ⇔ + = ⇔ + + = ⇔ − + = ⇔ − − = ( ) ( ) ( ) 2 2 2 log 2 log 2 2 0 2 1 0 x x x x x x x x x x ⇔ + = ⇔ + = ⇔ − − = ⇔ − + =
• 89. ( )) log 12 3. log 4 1 0x x e x + − + = 5 2 5 3 ) log log 2 0f x x− + = 16 4x x⇔ = − ∨ = 5 log 1 0x⇔ − = 5 5 2 log 2 0 log 2 5 25 x x x x − = = = = { }5,25HP = { }4HP = ( ) ( ) ( ) ( ) 3 3 1 2 log 12 log 4 1 log 12 log 4 log 12 1 64 12 64 0 16 4 0 x x x x x x x x x x x x x x − ⇔ + − = − ⇔ + − = + ⇔ = ⇔ + − = ⇔ + − = ( )( ) 5 2 5 5 5 log 3. log 2 0 log 1 log 2 0 x x x x ⇔ − + = ⇔ − − = 5 1 log 1 5 5 x x x ⇔ = ⇔ = ⇔ = V
• 90. ( ) ( ) 32 2 2 ) log 1 log 1 10 0g x x+ − + − = ( )2 log 1 5 0x⇔ + − = ( ) ( ) 2 2 2 log 1 2 0 log 1 2 1 2 1 1 4 3 4 x x x x x − ⇔ + + = ⇔ + = − ⇔ + = ⇔ = − ⇔ = − 3 ,31 4 HP   = −    V ( )2 5 log 1 5 1 2 32 1 31 x x x x ⇔ + = ⇔ + = ⇔ = − ⇔ = ( ) ( ) ( ){ } ( ){ } 2 2 2 2 2 log 1 3. log 1 10 0 log 1 5 log 1 2 0 x x x x ⇔ + − + − = ⇔ + − + + =

Jul 23, 2017

#### Good News 1968 (Vol XVII No 07-08) Jul-Aug

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