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Eksponensiële vergelykings

Module by: Free High School Science Texts Project. E-mail the author

Ekponensiële Vergelykings van die vorm ka(x+p)=mka(x+p)=m

Eksponensiële vergelykings het in die algemeen die veranderlike in die mag. Die volgende is voorbeelde van eksponensiële vergelykings:

2 x = 1 2 - x 3 x + 1 = 2 4 3 - 6 = 7 x + 2 2 x = 1 2 - x 3 x + 1 = 2 4 3 - 6 = 7 x + 2
(1)

Jy behoort reeds vertroud te wees met eksponensiële notasie. Oplos van eksponensiële vergelykings is eenvoudig, solank ons onthou hoe om die eksponentwette toe te pas.

Ondersoek: Oplos van Eksponensiële Vergelykings

Los die volgende vergelykings op deur die tabel te voltooi:

Table 1
2 x = 2 2 x = 2 x x
  -3 -2 -1 0 1 2 3
2 x 2 x              
Table 2
3 x = 9 3 x = 9 x x
  -3 -2 -1 0 1 2 3
3 x 3 x              
Table 3
2 x + 1 = 8 2 x + 1 = 8 x x
  -3 -2 -1 0 1 2 3
2 x + 1 2 x + 1              

Algebraïese Oplossing

Definition 1: Gelykheid van Eksponensiële Funksies

As aa 'n positiewe getal is so dat a>0a>0, (behalwe wanneer a=1a=1
) dan:

a x = a y a x = a y
(2)

as en slegs as:

x = y x = y
(3)
(As a=1a=1, dan kan xx en yy verskil.)

Dit beteken dat indien ons al die terme met dieselfde grondtal kan skryf, kan ons die eksponensiële vergelykings oplos deur die eksponente gelyk te stel. Neem byvoorbeeld die vergelyking 3x+1=93x+1=9. Dit kan geskryf word as:

3 x + 1 = 3 2 3 x + 1 = 3 2
(4)

Aangesien die grondtalle gelyk is (aan 3), weet ons dat die eksponente gelyk moet wees. Daarom kan ons skryf:

x + 1 = 2 x + 1 = 2
(5)

Dit gee:

x = 1 x = 1
(6)

Metode: Oplos van Eksponensiële Vergelykings

  1. Probeer om al die terme met dieselfde grondtal te skryf.
  2. Stel die eksponente van die grondtalle van die LK en RK gelyk.
  3. Los die vergelyking wat in die vorige stap verkry is, op.
  4. Bevestig die antwoorde.
Ondersoek : Eksponensiële Getalle

Skryf die volgende met dieselfde grondtal. Die grondtal is eerste in die lys. Byvoorbeeld in die lys 2, 4, 8, is die grondtal twee(2) en ons kan 4 skryf as 2222.

  1. 2,4,8,16,32,64,128,512,1024
  2. 3,9,27,81,243
  3. 5,25,125,625
  4. 13,169
  5. 2x2x, 4x24x2, 8x38x3, 49x849x8
Exercise 1: Oplos van Eksponensiële Vergelykings

Los op vir xx: 2x=22x=2

Solution
  1. Step 1. Probeer al die terme skryf met dieselfde grondtal :

    Al die terme word geskryf met dieselfde grondtal:

    2 x = 2 1 2 x = 2 1
    (7)
  2. Step 2. Stel die eksponente gelyk :
    x = 1 x = 1
    (8)
  3. Step 3. Toets jou antwoord :
    LK = 2 x = 2 1 = 2 RK = 2 1 = 2 = LK LK = 2 x = 2 1 = 2 RK = 2 1 = 2 = LK
    (9)

    Aangesien beide kante van die vergelyking dieselfde is, is die antwoord korrek.

  4. Step 4. Skryf die finale antwoord :
    x = 1 x = 1
    (10)

    is die oplossing van 2x=22x=2.

Exercise 2: Oplos van Eksponensiële Vergelykings

Los op:

2 x + 4 = 4 2 x 2 x + 4 = 4 2 x
(11)
Solution
  1. Step 1. Probeer om al die terme met dieselfde grondtal te skryf :
    2 x + 4 = 4 2 x 2 x + 4 = 2 2 ( 2 x ) 2 x + 4 = 2 4 x 2 x + 4 = 4 2 x 2 x + 4 = 2 2 ( 2 x ) 2 x + 4 = 2 4 x
    (12)
  2. Step 2. Stel die eksponenete gelyk :
    x + 4 = 4 x x + 4 = 4 x
    (13)
  3. Step 3. Los op vir xx :
    x + 4 = 4 x x - 4 x = - 4 - 3 x = - 4 x = - 4 - 3 x = 4 3 x + 4 = 4 x x - 4 x = - 4 - 3 x = - 4 x = - 4 - 3 x = 4 3
    (14)
  4. Step 4. Toets jou antwoord :
    LK = 2 x + 4 = 2 ( 4 3 + 4 ) = 2 16 3 = ( 2 16 ) 1 3 RK = 4 2 x = 4 2 ( 4 3 ) = 4 8 3 = ( 4 8 ) 1 3 = ( ( 2 2 ) 8 ) 1 3 = ( 2 16 ) 1 3 = LK LK = 2 x + 4 = 2 ( 4 3 + 4 ) = 2 16 3 = ( 2 16 ) 1 3 RK = 4 2 x = 4 2 ( 4 3 ) = 4 8 3 = ( 4 8 ) 1 3 = ( ( 2 2 ) 8 ) 1 3 = ( 2 16 ) 1 3 = LK
    (15)

    Aangesien beide kante dieselfde is, is die oplossing korrek.

  5. Step 5. Skryf die finale antwoord :
    x = 4 3 x = 4 3
    (16)

    is die oplossing van 2x+4=42x2x+4=42x.

Oplos van Eksponensiële Vergelykings
  1. Los die volgende eksponensiële vergelykings op:
    1. 2x+5=252x+5=25
    2. 32x+1=3332x+1=33
    3. 52x+2=5352x+2=53
    4. 65-x=61265-x=612
    5. 64x+1=162x+564x+1=162x+5
    6. 125x=5125x=5
    Kliek hier vir die oplossing
  2. Los op 39x-2=2739x-2=27
    Kliek hier vir die oplossing
  3. Los op vir kk: 81k+2=27k+481k+2=27k+4
    Kliek hier vir die oplossing
  4. Die groei van alge in 'n poel kan gemodelleer word met die funksie f(t)=2tf(t)=2t. Vind die waarde van tt sodat f(t)=128f(t)=128
    Kliek hier vir die oplossing
  5. Los op vir xx: 25(1-2x)=5425(1-2x)=54
    Kliek hier vir die oplossing
  6. Los op vir xx: 27x×9x-2=127x×9x-2=1
    Kliek hier vir die oplossing

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