Exponential Equations

Practice - Exponential equations Restriction k >0 Equ at io n Typ e 1

1. 2.

2

0,5 x ⋅ 2 2 x + 2 =

5. 6. 7. 8. 9. 10 .

x

5

1 =  5

0,125 ⋅ 4 3   4

x +1

−3 x − 4

2 x −3

 2  =    8 

−x

2

9  4x ⋅  = 16 3

D=R \ {0}

3 x − 4 ⋅ 27 3− 2 x = 9 3 x −3 9

0 , 25 x 2 − 2 x −8

3

−x 3

4

 3   5

12 . 13 . 14 .

2x

2

6 ⋅  5

6x −4

2− x

=

= 64

2 x − 4 ⋅ 8 3− 2 x = 4 3 x − 3

4

x +5

Z={-2,4}

-x+ =4

1 2

Z={- 3 }

-x2+3x+4=0

Z={-1,4}

3x=18

Z={6}

x2-x-2=0

Z={-1,2}

3 −5 x +5 = 36 x −6

Z={1} Z={-3,11}

1 2

Z={- 3 ,0, 3 } Z={2}

x2-5x+6=0

Z={2,3}

x2-5x+4=0

Z={1,4}

-11x=-11

Z={1}

x −1 2 = x +1 3

1−x

3

x2-2x-8=0

25 36

=1

 1  x +1 =   2

4 3

Z={ }

x -8x-33=0 -x3+3x=0

x 2 −5 x + 6

− 5 x +10

Answe r

2

= 3

 3  ⋅9 =   3   

5x   6

11 .

15 .

1 64

1 3 ⋅   = 81 3 x2

Hints

3x=4

3 5 x −8 = 9 x − 2

3. 4.

Su bst itu ti on

D=R \ {-1}

Z={5} Z={10}

x +17

32 x − 7 = 0,25 ⋅ 128 x −3 Typ e 2

16 . 17 . 18 .

3 x + 2 − 3 x = 72

3x=k

2 x +3 − 2 x = 112 2 x − 2 x −4 = 15

8k=72 7 ⋅ 2 x = 112

2x=k

k=16

k=9 2x=16

Z={2} Z={4} Z={4}

1

19 . 20 . 21 . 22 . 23 . 24 . 25 . 26 . 27 . 28 . 29 . 30 . 31 . 32 . 33 .

5 x + 3 ⋅ 5 x − 2 = 140

5x=k

k=125

Z={3}

3 2 x + 2 + 3 2 x = 30

32x=k

k=3

Z={ }

3 2 x −1 + 3 2 x − 2 − 3 2 x − 4 = 315

32x=k

35k=315*81 k=36

Z={3}

2 x + 2 − 2 x −1 = 14

2x=k

Z={2}

3 x +1 − 3 x − 3 x −1 = 15

3x=k

Z={2}

4

x−2 2

−2

x +1

x +1 3

=8

3 2x +

4 = 2 ⋅ 3 2 x +1 − 9 x 27

Z={2}

1 x ⋅ 2 − 4 ⋅ 2 x + 15 = 0 4

− 15

8 ⋅ 5 x + 7 ⋅ 5 x −1 = 22 + 5 x +1

1 2

5x=k

Z={1}

32x=k

7 ⋅ 4 x − 2 2 x +1 = 26 + 7 ⋅ 4 x −1

4x=k

3 ⋅ 2 x − 20 = 2 x −1

2x=k

1 2

Z={ − 1 } 1 2

Z={ 1 } Z={3} Z={1}

7 x + 2 + 2 ⋅ 7 x −1 = 345 2 x +1 + 3 ⋅ 2 x −1 − 5 ⋅ 2 x + 6 = 0

2x=k

5 ⋅ 2 3 x + 8 x − 2 ⋅ 41,5 x = 16

23x=k

2x 2

2

2

− x+2

+ 5 ⋅ 2x

x −3 x

1 − 5⋅  2

2+

− x −1

= 26

3− x +2 x

= 11

2x 2

2

2 3

Z={ }

=k

Z={-1,2}

=k

Z={-3}

−x

x −3 x

Z={2}

Typ e 3

34 . 35 . 36 . 37 . 38 . 39 . 40 .

k2+3k-108=0 k1=9 ; k2=-12

3 x +1 + 9 x = 108

3x = k

2 ⋅ 16 x − 17 ⋅ 4 x + 8 = 0

4x=k

4x − 9 ⋅ 2x + 8 = 0

2x=k

k2-9k+8=0 k1=8 ; k2=1

Z={0,3}

4x − 8⋅ 2x = 0

2x=k

k2-8k=0

Z={3}

2 x +1 + 4 x = 80

2x=k

k2+2k-80=0 k1=-10 ; k2=8

Z={3}

4x + 8 = 6 ⋅ 2x

2x=k

k2-6k+8=0 k1=2 ; k2=4

Z={1,2}

4 x − 10 ⋅ 2 x −1 = 24

2x=k

k2-5k-24=0 k1=-3 ; k2=8

Z={3}

2k2-17k+8=0 k1=8 ; k2=

Z={2} 1 2

3 2

1 2

Z={ ,− }

2

41 4x 5⋅ 2x − =8 . 2 42 8 ⋅ 3 x − 2 + 1 = 9 x −1 . 43 9x-2·3 x −1 -7=0 . 44 9 x − 486 = 3 x + 2 . 45 4 2 x +1 = 65 ⋅ 4 x −1 − 1 . 46 2 2 x + 2 − 3 ⋅ 2 x + 2 + 8 = 0 . 47 3 x + 2 + 9 x +1 = 810 . 48 25 x − 5 x +1 + 5 = 5 x . 49 3 2 x −1 + 3 x +1 = 12 . 50 5 2 x −1 + 5 x +1 = 250 . x x 51 2 2 3 ⋅ 4 − 7 ⋅ 2 = 20 . 2 52  1  3 x − x   + 3⋅8 3 − 4 = 0 . 2

2x=k

k3+3k2-4=0; k1=1 ; k2=-2

53 . 54 . 55 .

  30 − 5 x = 5 − x +3

1   =k 2

5x=k

-k2+30k-125=0 k2=5

9

3x=k

3x + 1   5

3

2x

x +1

=

3x=k 3k2-2k-21=0 ; k1=-

7 3

; k2=3

Z={3}

4x=k

Z={-2,1}

2x=k

Z={1,0}

3x=k

k2+k-90=0 k1=-10 ; k2=9

Z={2}

5x=k

k2-6k+5=0 k1=1 ; k2=5

Z={0,1}

3x=k

k1=-12 ; k2=3

Z={1}

5x=k

k2+25k-1250=0 k2=-50

x

k1=-

x

k1=25 ; Z={2} Z={4}

5 k2=4 3

3k2-28k+3=0 k1=

Z={0}

k1=25 ; Z={2,1} 1 ; k2=9 3

Z={-1,2} Z={-2}

x

=0

Z={1}

3x=k

22 = k

−2

Z={1,3} Z={2}

3x=k

28 3

1 − 24 ⋅ 5 − x −   5

k2-10k+16=0 k1=2 k2=8

1   =k 5

Typ e 4

56 11x −7 = 17 7 − x . 57 7 ⋅ 3 x +1 − 5 x + 2 = 3 x + 4 − 5 x +3 .

11x −7 =

1 17 x −7

(11 ⋅ 17 ) x −7

=1

Z={7}

 3   =k 5

Z={-1}

2x=k

Z={0,2}

x

Bo nus

58 . 59 . 60 .

1 + 21− x = 1 x 2 −2

(2 + 3 ) + (2 − 3 ) x

4

3x

−7⋅4

2x

x

=4

+ 14 ⋅ 4 − 8 = 0 x

(2 + 3 ) x

4 =k

x

=k

k2-4k+1=0 k1= 2 − 3

k2= Z={1,-1}

2+ 3 1 2

Z={ 0, ,1 }

3

61 .

4 sin

62 . 63 .

2 2 sin x cos x = 2

2

1   2

x

= 2 ⋅ 2 cos x

sin2 x

=

1 2

dla cosx=-1

wynik: x1= ∏ +2k ∏ ;

1 2 1 sin2x= 2

∏ ∏ + 2k ∏ ; x3= − + 2k ∏ 3 3 ∏ 5 x1= + k ∏ ; x2= ∏ +k ∏ 12 12

dla cosx=

wynik: x2=

dla

wynik:

dla sinx= dla sinx=-

2 2 2 2

wynik: x1=

∏ 3∏ + 2k ∏ ;x2= + 2k ∏ ; 4 4

wynik: x3=-

∏ 5 + 2k ∏ ;x4= ∏ +2k ∏ 4 4

4