10 – Exponentiële en logaritmische functies
10.4 – Rekenregels voor logaritmen
10 Gebruik dat g log a
g
= a en
a
b
3
7
3
log 5
7
log11
c
 2
 
 3
2
3
g
( )
log g a = a
=5
= 11
log 2
=2
( )
log (10 ) = 1,3
d 4 log 43 = 3
e
10
f
1
2
  1 3 
log     = 3
 2  


g
10
log1000 = 10 log 103 = 3
h
0,5
i
2
log16 = 2 log(24 ) = 4
j
x
log( x 2 ) = 2
p
log( p ) =
k
l
11 a
b
1,3
( )
log ( 0, 25) =
0,5
(
log ( 0,5 )
1
a
2
5
a
log13
p
log( p 2 ) =
2
)=2
1
2
= 13
log a + 2 log b = 2 log ( ab )
x
log x − 5 log y = 5 log  
 y
( )
c 6 ⋅ 3 log a = 3 log a 6
d
5
1

1
log 2 + 5 log xy + 5 log   = 5 log  2 ⋅ xy ⋅  = 5 log ( 2 x )
y
 y

)
log ( x ⋅ xy ) =
e
3
log a + 3 log ab = 3 log
f
7
log x 2 − 7 log ( xy ) =
g
2
( )
7
(
( )
log ( x y )
a ab = 3 log a b
2
7
3
 x ⋅ xy  2
 x 
log x − 2 log y + 2 log xy = 2 log 
 = log 


y


 y
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10 – Exponentiële en logaritmische functies
h
2

 1
x  2
= log 
log x − 2 log y − 2 log xy = 2 log 

 y ⋅ xy 
y y



i
 x3 
3 ⋅ 2 log x − 1 = 2 log x3 − 1 = 2 log x 3 − 2 log 2 = 2 log  
 2
 
j
2 ⋅ 3 log x + 3 ⋅ 3 log ( 2 y ) = 3 log 

( )
= 3 log x + 3
( )
( x)
log (8 y ) = log ( 8 xy )
3
3
(
2
3
 + log ( 2 y )

)
log ( ab ) = 2 log a + 2 log b
2
b
3
c
7
log a 2 b = 7 log a 2 + 7 log b = 2 ⋅ 7 log a + 7 log b
d
1
2
1
1
1
 2 xy 2  12
log 
 = log 2 + 2 log x + 2 ⋅ 2 log y − 2 log z
 z 
5
3
3
12 a
e



a
log   = 3 log a − 3 log b
b
( )
( )
 1 5
1
3
log x x = log x + log  x 2  = log x + ⋅ 5 log x = ⋅ 5 log x
 
2
2
 
(
)
5
5
f
3
log ( abcd ) = 3 log a + 3 log b + 3 log c + 3 log d
g
2
3
2
2
2
2
2
 1
 a 2 b  32
1
2
log 
 = log a + 3 log b − 3 log  c 2  = 2 ⋅ 3 log a + 3 log b − ⋅ 3 log c
 
2
 c
 
( )
(
)
( )
( )
h 2 log 3x −1 y 3 = 2 log 3 + 2 log x −1 + 2 log y 3 = 2 log 3 − 2 log x + 3 ⋅ 2 log y
i
j
2
2

1
log  x −3 ⋅  = 2 log x −3 + 2 log y −1 = −3 ⋅ 2 log x − 2 log y
y

( )
log
(
3
ab
2
)

= log  ab 2


2
( )
( )
1
3
 1 2
1
2
 = ⋅ log ab2 = ⋅ 2 log a + 2 log b
 3
3
3

( )
k 6 log(2ab) = 6 log(2) + 6 log( a) + 6 log(b)
l
4
log(4a 2 bc3 ) = 4 log(4) + 4 log( a 2 ) + 4 log(b) + 4 log(c3 ) =
1 + 2 ⋅ 4 log(a ) + 4 log(b) + 3 ⋅ 4 log(c)
m 3 log( x 2 y −3 ) = 3 log( x 2 ) + 3 log( y −3 ) = 2 ⋅ 3 log( x ) − 3 ⋅ 3 log( y )
n
4
log(
x2
y3
) = 4 log( x 2 ) − 4 log( y 3 ) = 2 ⋅ 4 log( x ) − 3 ⋅ 4 log( y )
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10 – Exponentiële en logaritmische functies
o
13 a
1

3 2
log( a ab ) = log(a ) + log( ab ) = 2 ⋅ log( a) + log  ( ab )  =




1
1
1
1
2 ⋅ 2 log( a) + ⋅ 2 log( a) + ⋅ 2 log(b3 ) = 2 ⋅ 2 log(a ) + 1 ⋅ 2 log(b)
2
2
2
2
2
2
3
log 5 ≈ 1, 46
b
10
c
1
3
3
2
2
2
3
2
2
log 2 ≈ 0,30
log 25 ≈ −2,93
d 3 log100 ≈ 4,19
e
f
g
3
log
99
100
1
≈ −4,19
100
log 2 ≈ −68,97
1
125
2
2
log(25) ≈~ 0,67 (exact : , want 1253 = 5 en 52 = 25, dus 125 3 = 25
3
h
50
log(33) ≈ 0,89
i
33
log(50) ≈ 1,12
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Practice 4-4 Level B

Extra Oefening Hs1 Exponentiële en Logaritmische Functies