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Ekponensiële vergelykings van die vorm k a ( 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

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:

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

Algebraïese oplossing

Gelykheid van Eksponensiële Funksies

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

a x = a y

as en slegs as:

x = y
(As a = 1 , dan kan x en y 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 3 x + 1 = 9 . Dit kan geskryf word as:

3 x + 1 = 3 2

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

x + 1 = 2

Dit gee:

x = 1

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 2 2 .

  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. 2 x , 4 x 2 , 8 x 3 , 49 x 8

Los op vir x : 2 x = 2

  1. Al die terme word geskryf met dieselfde grondtal:

    2 x = 2 1
  2. x = 1
  3. LK = 2 x = 2 1 = 2 RK = 2 1 = 2 = LK

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

  4. x = 1

    is die oplossing van 2 x = 2 .

Los op:

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

    Aangesien beide kante dieselfde is, is die oplossing korrek.

  5. x = 4 3

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

Oplos van eksponensiële vergelykings

  1. Los die volgende eksponensiële vergelykings op:
    1. 2 x + 5 = 2 5
    2. 3 2 x + 1 = 3 3
    3. 5 2 x + 2 = 5 3
    4. 6 5 - x = 6 12
    5. 64 x + 1 = 16 2 x + 5
    6. 125 x = 5
  2. Los op 3 9 x - 2 = 27
  3. Los op vir k : 81 k + 2 = 27 k + 4
  4. Die groei van alge in 'n poel kan gemodelleer word met die funksie f ( t ) = 2 t . Vind die waarde van t sodat f ( t ) = 128
  5. Los op vir x : 25 ( 1 - 2 x ) = 5 4
  6. Los op vir x : 27 x × 9 x - 2 = 1

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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