Vergelijkingen oplossen
1
a

x = 2 3 = 8

x = 7

x = 4

x = 3 1 2

x = 64

x = 3 of x = 3

x = 1 8

x = 4 of x = 2

b

log ( x ) = 3 log ( 6 )

3 log ( x ) = log ( 6 )

log ( x ) = log ( 6 3 )

log ( x 3 ) = log ( 6 )

x = 216

x 3 = 6

x = 6 3


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

log ( 6 ) + log ( 1 x ) = log ( x )

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

log ( 6 x ) = log ( x )

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

6 x = x

x = 2 ‐3 = 1 8

x = 6 (Niet x = 6 , want x > 0 )

c

2 log ( x ) + 2 log ( 8 ) = 2 log ( 12 )

2 log ( 95 ) 2 log ( x ) = 2 log ( 5 )

2 log ( 8 x ) = 2 log ( 12 )

2 log ( 95 x ) = 2 log ( 5 )

8 x = 12 x = 1 1 2

95 x = 5 x = 19


5 log ( x ) 5 log ( 2 ) = 5 log ( 7 )

log ( x ) + log ( 40 ) = 4

5 log ( x 2 ) = 5 log ( 7 )

log ( 40 x ) = 4

x 2 = 7 x = 14

40 x = 10 4 x = 250


log ( x ) log ( 5 ) = log ( 70 )

2 log ( x ) 2 log ( 3 ) = 2 log ( 12 ) 2 log ( x )

log ( x 5 ) = log ( 70 )

2 log ( x 3 ) = 2 log ( 12 x )

x = 350

x 3 = 12 x x 2 = 36

Alleen x = 6 voldoet.

2
a

Domein f : alle getallen x met x 0
Domein g : de getallen x met x > 0
Domein h : de getallen x met x > 6

b

Regel 3, maar die geldt alleen voor positieve x , dus het deel rechts van de y -as.

c

2 log ( x 2 ) = 2 log ( 1 2 x + 3 ) , dus x 2 = 1 2 x + 3
2 x 2 x 6 = 0 , dus x = 2 of x = 1 1 2 .

d

Omdat je de vergelijking 2 log ( x 2 ) = 2 log ( 1 2 x + 3 ) oplost. En alleen voor positieve x geldt: 2 log ( x ) = log ( x 2 ) !

3
  • 2 log ( x 2 ) = 1 + log ( 1 2 x + 4 )
    log ( x 2 ) 2 = log ( 10 ) + log ( 1 2 x + 4 )
    ( x 2 ) 2 = 10 ( 1 2 x + 4 )
    x 2 9 x 36 = 0
    x = 12 of x = 3
    Alleen x = 12 is oplossing.

  • 2 log ( x + 1 ) = 2 + log ( x 8 )
    log ( x + 1 ) 2 = log ( 100 ( x 8 ) )
    ( x + 1 ) 2 100 ( x 8 ) = 0
    x = 9 of x = 89 , beide voldoen.

  • log ( x + 2 ) log ( x 7 ) = log ( 2 x 6 )
    log ( x + 2 x 7 ) = log ( 2 x 6 )
    x + 2 x 7 = 2 x 6
    2 x 2 21 x + 40 = 0
    x = 2 1 2 of x = 8 .
    Alleen x = 8 is oplossing.

Gemengde opgaven
4
a

Invullen geeft 25  km/u.

b

Als s groter wordt, wordt de 2 log ( ... ) groter en dus 20 2 log ( ... ) kleiner.

c

Als je 2 s in de formule v = 20 2 log ( s 10.000 ) invult in plaats van s , krijg je: v = 20 2 log ( 2 s 10.000 ) = 20 ( 2 log ( 2 ) + 2 log ( s 10.000 ) ) , dus v wordt 1 kleiner want 2 log ( 2 ) = 1 .

d

v = 20 2 log ( s 10.000 )
v = 2 log ( 2 20 ) 2 log ( s 10.000 ) = 2 log ( 10.000 s ) 2 20 , dus:
2 v = 2 20 10.000 s , dus 2 v s = 2 20 10.000 10 10 , want 2 10 1000 = 10 3 .

5
a

Bij de gehoorgrens hoort geluidsdrukniveau L = 10 log ( I 0 I 0 ) = 0 , want log ( 1 ) = 0 .
Bij de pijngrens hoort geluidsdrukniveau 10 log ( 10 10 12 ) = 10 log ( 10 13 ) = 130 .

b

10 log ( 2 I I 0 ) = 10 log ( 2 ) + 10 log ( I I 0 ) = 10 log ( 2 ) + 80 = 83,0

c

Noem de gezochte afstand x , dan:
74 = L 0 10 log ( 2 π x ) en
77 = L 0 10 log ( 40 π ) , dus (trek de twee vergelijkingen van elkaar af en deel door 10 )
log ( 2 π x ) = log ( 40 π ) + 0,3 = log ( 40 π ) 10 0,3 , dus x = 20 10 0,3 40 meter.

6

log ( 1 5 ) = log ( 2 ) log ( 10 ) = p 1

log ( 3200 ) = log ( 32 ) + log ( 100 ) = 5 p + 2

2 log ( 10 ) = log ( 10 ) log ( 2 ) = 1 p

5 log ( 10 ) = log ( 10 ) log ( 5 ) = log ( 10 ) log ( 10 ) log ( 2 ) = 1 1 p