ЕКСПОНЕНЦИЈАЛНИ РАВЕНКИ-РЕШЕНИ ЗАДАЧИ 1. Провери дали дадениот број е решение на експоненцијалната равенка: а) 0x = , 4 78x x= ; б) 1x = − , 13 1x+ = ; в) 2x = , 17 3 7 28x x−− ⋅ = ; г) 3x = − , 6 2 4 8 400x x⋅ + ⋅ = ; д) 1x = , 3 3 5 356 2 3 0x x− −+ − = ; ѓ) 1x = , 2 8 15 1x⋅ − = − . Реши ги равенките: 2. а) 2 16x = ; б) 4 16x = ; в) 27 49x = ; г) 2 18 64x− = ; д) 8 64x = ; ѓ) 53 27x− = ; е) 32 16x− = . 3. а) 33 6 108x⋅ = ; б) 12 8 16x+⋅ = ; в) 0,2 5x = ;

г) 20,5 4x = ;

д) 0,1 0,001x = ;

ѓ) 29 3x+ = .

4. а) 2 33 2

x =

;

б) 2 1 74 5

5 4

x x− + =

;

в) 4 53 25

5 9

x− =

;

г) 2 2 12 49

7 4

x x− + =

;

д) 1 449 4

16 7

x+ =

;

ѓ) 4 135 36

6 25

x − =

;

е) ( )12

3 44,581

x =

.

5. а) 2 34 128x = ;

б) 21 1

64 8

x− =

;

в) ( ) 11 424 2

x x− −= ;

г)

5 432 8 9

3 27 4x

− ⋅ =

;

д) 1

3 310 100x −

= ;

ѓ) 12 2 2x− = .

6. а) 24 3 4 13x x−− ⋅ = ; б) 2 12 2 2 104x x x− ++ + = ; в) 1 1 2 1 23 3 3 5 5 5x x x x x x+ − − − −+ + = + + ; г) 3 4 221 3 5 3 5x x x x+ + +⋅ + = + ; д) 1 52 5 2 7 2x x− −+ ⋅ = ⋅ ; ѓ) 1 37 2 5 2 468x x+ −⋅ + ⋅ = . 7. а) 4 9 2 8 0x x− ⋅ + = ; б) 16 12 4 32 0x x− ⋅ + = ; в) 9 27 12 3x x+ = ⋅ ; г) 2 8 2 7x x−− ⋅ = ; д) 1 149 55 7 56 0x x+ ++ ⋅ − = ; ѓ) 125 49 5 2 0x x+ + ⋅ − = .

РЕШЕНИЈА 1. а) Бројот 0x = е решение на равенката 4 78x x= , затоа што 0 04 78 1 1= ⇔ = ; б) Бројот 1x = − е решение на равенката 13 1x+ = , затоа што 1 1 03 1 3 1 1 1− + = ⇔ = ⇔ = ; в) Бројот 2x = е решение на равенката 17 3 7 28x x−− ⋅ = , затоа што

2 2 17 3 7 28 49 3 7 28 28 28−− ⋅ = ⇔ − ⋅ = ⇔ = ; г) Бројот 3x = − не е решение на равенката 6 2 4 8 400x x⋅ + ⋅ = , затоа што

3 33 3

6 4 3 1 3 32 1 96 16 2 4 8 400 400 400 400 4002 8 4 128 128 128

− − ⋅ + +⋅ + ⋅ = ⇔ + = ⇔ + = ⇔ = ⇔ ≠ ;

д) да; ѓ) не. 2. а) 42 16 2 2 4x x x= ⇔ = ⇔ = ; б) 24 16 4 4 2x x x= ⇔ = ⇔ = ; в) 2 2 27 49 7 7 2 2 1x x x x= ⇔ = ⇔ = ⇔ = ;

г) 2 1 2 1 2 38 64 8 8 2 1 2 2 32

x x x x x− −= ⇔ = ⇔ − = ⇔ = ⇔ = ;

д) 2x = ; ѓ) 8x = ; е) 1x = − . 3. а) 33 6 108/ :3x⋅ =

3 3 2 36 36 6 6 3 22

x x x x= ⇔ = ⇔ = ⇔ = ;

б) 12 8 16/ : 2x+⋅ = 18 8 1 1 0x x x+ = ⇔ + = ⇔ = ;

в) ( )12 10,2 5 5 5 5 5 5 5 1 110 5

x xxx x x x− − = ⇔ = ⇔ = ⇔ = ⇔ = ⇔ − = ⇔ = −

;

г) 2 2

2 2 25 10,5 4 4 4 2 2 110 2

x xx x x− = ⇔ = ⇔ = ⇔ = ⇔ = −

;

д) 3x = ; ѓ) 32

x = − .

4. а) ЗАПОМНИ 13 1 2

22 33

− = =

!

Според тоа 12 3 2 2 1

3 2 3 3

x x

x−

= ⇔ = ⇔ = −

;

б) ( )72 1 7 2 1 14 5 4 4 2 1 7 3 6 2

5 4 5 5

xx x x

x x x x+− + − − = ⇔ = ⇔ − = − + ⇔ = − ⇔ = −

;

в) 4 5 4 5 23 25 3 3 34 5 2

5 9 5 5 4

x x

x x− − −

= ⇔ = ⇔ − = − ⇔ =

;

г) ( )2 2 1 2 2 2 12 49 2 2 2 4 2 5 0 0

7 4 7 7

x x x x

x x x x− + − − +

= ⇔ = ⇔ − = − − ⇔ = ⇔ =

;

д) 3x = − ; ѓ) 132

x = ; е) 8x = − .

5. а) Бидејќи 2 5 7128 4 32 2 2 2= ⋅ = ⋅ = , се добива 2 2 7334 128 4 2x x= ⇔ =

ЗАПОМНИ 7

73 32 2= !

Според тоа 7 7

2 2 23 3

77 734 2 2 2 43 4 12

x x x x x⋅= ⇔ = ⇔ = ⇔ = ⇔ = ;

б) ( )2 2 3

26 3 12 26 3

1 1 1 1 3 12 2 2 2 1264 8 2 2 2 8

x xx x x x

− −−−− − = ⇔ = ⇔ = ⇔ = ⇔ = − ⇔ = −

;

в) ( )

( )

( )12 121 11 24 22 4 1 1 34 2 2 2 1 0

2 4 4

x xx x

x x x x

− − − −

= ⇔ = ⇔ − − = ⇔ − − =

1/2

31 1 4 1 242 2

x± + ⋅ ±= = , па решенија на равенката се 1 2

3 1,2 2

x x= = − ;

г)

( )

155 3 44 3 2

33 2

5 1 53 2 4 83

2 8 9 2 2 33 27 4 3 3 2

2 2 3 2 2 2 5 5 51 8 73 3 2 3 3 3 7

xx

x xx

x x

−−

⋅ ⋅ ⋅ −

⋅ = ⇔ ⋅ = ⇔

⇔ ⋅ = ⇔ ⋅ = ⇔ + = ⇔ = ⇔ =

д) 1x = ; ѓ) 52

x = (Упатство: 1 1 311 2 2 22 2 2 2 2 2

+= ⋅ = = ).

6. а)

2 22

3 134 3 4 13 4 3 4 4 13 4 1 13 4 13 4 16 24 16

x x x x x x x x− − − ⋅ = ⇔ − ⋅ ⋅ = ⇔ − = ⇔ ⋅ = ⇔ = ⇔ =

б)

2 1 22 2 2 104 2 2 2 2 2 104

1 13 1042 1 2 104 2 104 24 4

x x x x x x

x x x

− + −+ + = ⇔ ⋅ + + ⋅ = ⇔

⇔ + + = ⇔ ⋅ = ⇔ =

84

13⋅

1 2 32 5x x⇔ = ⇔ =

в) 1 1 2 1 2 1 1 1 13 3 3 5 5 5 3 3 5 13 9 5 25

x x x x x x x x+ − − − − + + = + + ⇔ + + = + + ⇔

2

31 31 1 1 93 5 / : 31 3 5 / 9 3 5 / :59 25 9 25 25

3 9 3 3 25 25 5 5

x x x x x x x

xx

x x

⇔ ⋅ = ⋅ ⇔ ⋅ = ⋅ ⋅ ⇔ = ⋅ ⇔

⇔ = ⇔ = ⇔ =

г) ( ) ( )3 4 2 4 2 321 3 5 3 5 21 3 3 5 5 3 21 81 5 25 125x x x x x x x x x x+ + + + + +⋅ + = + ⇔ ⋅ − = − ⇔ − = − ⇔

( ) ( ) ( ) 1003 60 5 100 / : 60 3 5x x x x⇔ ⋅ − = ⋅ − − ⇔ = ⋅5

603

3 5 15 3

x

x ⇔ = ⇔ = −

;

д) 4x = − ; ѓ) 5x = .

7. а) ( ) ( )2

1/2

9 81 32 9 74 9 2 8 0 2 9 2 8 0 22 2

x x x x x ± − ±− ⋅ + = ⇔ − ⋅ + = ⇔ = =

( ) 112 8 3x x= ⇔ =

( ) 222 1 0x x= ⇔ = ;

б) ( )216 12 4 32 0 4 12 4 32 0x x x x− ⋅ + = ⇔ − ⋅ + =

( )1/2

12 144 128 12 442 2

x ± − ±= =

( ) 2 311

34 8 2 2 2 32

x x x x= ⇔ = ⇔ = ⇔ =

( ) 224 4 1x x= ⇔ = ;

в) ( )29 27 12 3 3 12 3 27 0x x x x+ = ⋅ ⇔ − ⋅ + =

( )1/2

12 144 108 12 632 2

x ± − ±= =

( ) 113 9 2x x= ⇔ =

( ) 223 3 1x x= ⇔ = ;

г) ( ) ( )2 282 8 2 7 2 7 / 2 2 8 7 2 2 7 2 8 02

x x x x x x x xx

−− ⋅ = ⇔ − = ⋅ ⇔ − = ⋅ ⇔ − ⋅ − =

( )1/2

7 49 32 7 922 2

x ± + ±= =

( ) 112 8 3x x= ⇔ =

( ) 222 1x x= − ⇔ не постои;

д) 1 21,x x= − не постои (Упатство: ( )21 149 7x x+ += ); ѓ) 1 2x = − , 2x не постои.