Diagnostische Toets Wis B VWO Hoofdstuk 5 Exponenten en logaritmen Getal en Ruimte wiskunde uitwerkingen www.uitwerkingensite.nl
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5.5
Diagnostische toets
bladzijde 48
1
a y = 3x
↓ translatie (2, -3)
f (x) = 3x-2 - 3
y=
()
1
3
x
↓ verm. x-as, 4
y = 4⋅
()
1
3
x
↓ translatie (0, - 6)
()
g x = 4⋅
()
1
3
x
-6
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b Voer in y1 = 3x-2 - 3 en y2 = 4 ⋅
()
1
3
x
- 6.
x
-1
0
1
2
3
4
f (x)
-3,0
-2,9
-2,7
-2
0
6
g(x)
6
-2
- 4,7
-5,6
-5,9
- 6,0
y
y=6
6
5
4
3
ƒ
2
g
1
0,22
–1
O
1
2
3
x
–1
–2
y = –3
–3
–4
–5
y = –6
–6
c Bf = 〈-3, →〉 en Bg = 〈- 6, →〉
d De optie intersect geeft x ≈ 0,22.
f (x) ≥ g(x) geeft x ≥ 0,22.
( ) -6 = 6
4 ⋅ ( ) = 12
( ) =3
( ) =( )
e g(x) = 6 geeft 4 ⋅
1
3
1
3
1
3
1
3
x
x
x
x
1
3
-1
x = -1
g(x) ≤ 6 geeft x ≥ -1
f f (4) = 6
Voor x ≤ 4 is -3 < f (x) ≤ 6.
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2
c 2 ⋅ 42 x -1 - 3 = 6611
a 5x−1 = 125⋅ 3
5
1
3
2 ⋅ 42 x -1 = 64
64
5x-1 = 53 ⋅ 5
1
3 3
42 x -1 = 32
32
5x-1 = 5
(2 )
x - 1 = 3 1
3
2
1
27
=
1
2 x-5
3
33
3
1
⋅ 32
1
3
2 x-5 = 3-3 ⋅ 32
d
1
3
2 x-5 = 3-2
2
( )
= 33
x+1
3
=3
2x - 2 = 3x + 3
-x = 5
x = -5 b 2x+2 + 2x-1 = 36
2x ⋅ 22 + 2x ⋅ 2-1 = 36
2x-2
3x+3
=
=
1
8
()
1
2
()
=( )
d 2x =
2
2x
3
2
1
8
x
1
x
2
3
( )
x
x
4 ⋅ 2 x + 12 ⋅ 2 x
= 36
2 x = 2 -3
4 12 ⋅ 2 x
= 36
2 x = 22
-33xx
x2 = -33x
x2 + 3x = 0
x(x + 3) = 0
x=0 ∨ x=-33
2
2
2x = 8
2x = 23
x = 3
4
3 x +1
c 3x + 1 = 3x + 54
54
3 ⋅ 3x = 3x + 54
54
2 ⋅ 3x = 54
54
3x = 27
27
3x
= 33
x = 3
a 9x-1 = 27x+1
2
3 x +11
+ 6 = 66
8
81
3x + 1 = 3 3x = 2 x = 23
x = 1 1
4
(3 )
3 x +11
1
2
1
2
2x =
2 1
2
x-1
( )
()
()
1
2
2x - 5 = -2 1
2
3
= 2
25
24 x - 2 = 22
5
4x - 2 = 5 4x = 7 x = 1 43
x = 4 1
3
b 3
2 x−5
=
2 x -1
a Stel H = b ⋅ gt.
7
gdag = 1+ 100
= 1, 07
Op t = 0 is H = 20, dus b = 20.
Dus H = 20 ⋅ 1,07t.
b Voer in y1 = 20 ⋅ 1,07x en y2 = 55.
De optie intersect geeft x ≈ 14,95.
Hierbij hoort 15 mei.
c
t
H
18
67,60
19
72,33
20
77,39
+ 4,73
+ 5,06
Dus op 20 mei.
5
a gdag = 1,10, dus gweek = 1,107 ≈ 1,949.
Het groeipercentage per week is 94,9.
1
b g8 uur = 1,10 3
≈ 1,
032
Het groeipercentage per 8 uur is 3,2.
6
1
a gjaar = 0,64, dus gmaand = 0, 6412
≈ 0, 963
De afname per maand is 3,7%.
b g5jaar = 0,645 ≈ 0,107.
De afname per 5 jaar is 89,3%.
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7
1200
= 0, 8.
1500
g3 dagen =
1
gdag = 0, 8 3
≈ 0, 928.
N = b ⋅ 0,928t
voor t = 4 is N = 1500
}
1500 = b ⋅ 0,9284
1500
=b
0, 9284
b ≈ 2020
Dus N = 2020 ⋅ 0,928t.
( ) ( ) ( )
log ( ⋅ 2 ) = log ( 2 ⋅ 2 ) = log ( 2 ) = −3
1
1
8 a 3 log 3 3 = 3log 31 ⋅ 32 = 3log 3 2 = 1 12
b
c
2
1
3
1
16
log 

d 2 log
−4
2
3
1
1
3
2
−3 3
2
()
1
3
 = 0, 6

1
1
⎛ 1
⎞
1
−
8 = 2 log ⎜ 2 ⋅ 23 ⎟ = 2 log 2 −2 ⋅ 2 2 = 2 log 2 2 = − 12
⎝2
⎠
0 ,6
)
(
( )
1
4
2
3
( )
bladzijde 49
9 a 4log(2x - 3) = 2
2x - 3 = 42
2x - 3 = 16
2x = 19
x = 9 12
b 3 + 3log(x) = 7
3
log(x) = 4
x = 34
x = 81
c
1
2
log(xx - 33)) = - 4
log(
4
x-3=
()
1
2
-4
( )
x - 3 = 2 -1
-4
x - 3 = 24
x - 3 = 16
x = 19
d 5 + 3 ⋅ 2log(
log(xx) = 20
3 ⋅ 2log(
log(xx) = 15
2
log(xx) = 5
log(
x = 25
x = 32
log(20)
≈ 21, 61
log( 2)
6
6
b 3
=
≈ 1, 94
log(30)  log(30)
 log(3) 

a 2x + 5 > 0 geeft 2x > -5
x > -2 12
10 a 5⋅ 2 log(20) = 5⋅
11
Dus Df = 〈-2 12 , →〉 en de verticale asymptoot is de lijn x = -2 12 .
x
-2
-1
0
1
2
3
6
f (x)
3
1,4
0,7
0,2
- 0,2
- 0,5
-1,1
y
3
ƒ
2
1
x=
1
–22
–2
–1
O
1
3
4
5
x
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b f (x) = -2 geeft 3 - 2log(2x + 5) = -2
-2log(2x + 5) = -5
2
log(2x + 5) = 5
2x + 5 = 25
2x + 5 = 32
2x = 27
x = 13 12
f (x) ≥ -2 geeft -2 12 < x ≤ 13 12 .
c f 5 1
2 = 3 - 2log(11 + 5) = 3 - 2log(16) = 3 - 4 = -1
Voor x ≤ 5 12 is f (x) ≥ -1.
( )
12 a Df = 〈-1, →〉 en de verticale asymptoot is de lijn x = -1.
-x + 6 > 0 geeft -x > - 6
x<6
Dg = 〈←, 6〉 en de verticale asymptoot is de lijn x = 6.
x
- 43
- 12
0
1
3
7
f (x)
2
1
0
-1
-2
-3
x
-2
2
4
5
5 12
5 43
5 78
g(x)
3
2
1
0
-1
-2
-3
y
4
3
g
2
1
–2
O
–1
1
2
3
4
5
6
x
–1
ƒ
–2
–3
x = –1
–4
x=6
1
b f (x) = 4 geeft 2 log(x + 1) = 4
x +1 =
()
1
2
4
x + 1 = 161
15
x = - 16
c g(-2) = log(2 + 6) = 2log(8) = 3
Voor x ≥ -2 is g(x) ≤ 3.
2
1
d f (x) = 2 geeft 2 log(x + 1) = 2
x +1 =
()
x +1 =
1
4
x=-
1
2
2
3
4
1
f (x) = -2 geeft 2 log(x + 1) = -2
x +1 =
()
1
2
-2
x + 1 = 4
x = 3
-2 ≤ f (x) ≤ 2 geeft - 43 ≤ x ≤ 3
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e Voer in y1 =
(
)
log x + 1
log
()
1
2
en y2 =
(
log - x + 6
()
log 2
).
De optie intersect geeft x ≈ - 0,85 en x ≈ 5,85
f (x) ≤ g(x) geeft - 0,85 ≤ x ≤ 5,85.
1
f f (x) = 1 geeft 2
log(x + 1) = 1
x +1 =
1
2
x = - 12
( )
g (x) = 1 geeft 2log(-x + 6) = 1
-x + 6 = 2
-x = - 4
x=4
Dus AB = 4 - - 12 = 4 12 .
I 
13 LJan = 78 geeft 10 log  Jan  = 78
 I 0

 I 
log  Jan  = 7, 8
 I0 
I Jan
= 107 ,8
10 -12
IJan = 10- 4,2 W/m2
I 
LFrits
= 80 geeft 10 log  Frits 
= 80
 I0 
I 
log  Frits  = 8
 I0 
I Frits
= 108
10 -12
IFrits = 10- 4 W/m2
I

LGerrit
= 81 geeft 10 log  Gerrit 
= 81
 I0 
I

log  Gerrit  = 8,1
I
 0 
I Gerrit
= 108,1
10 -12
IGerrit = 10-3,9 W/m2
Itotaal = 10- 4,2 + 10- 4 + 10-3,9 W/m2
 10 - 4,2 + 10 - 4 + 10 -3,9 
Ltotaal = 10 log 
 ≈ 85 dB
10 -12

Het geluidsniveau van de drie brommers samen is 85 dB.
Exponentenenlogaritmen37
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N
103
2
3
4
5
6 7 8 9104
2
3
4
5
6 7 8 9105
14 a
1990
1993
1995
1999
2001
2004
2005
2005
jaar
jaar
b Rechte lijn op logaritmisch papier dus N = b ⋅ g .
t
Lijn door (0, 15 000) en (15, 4300), dus g15 jaar =
1
15
 4300 
gjaar = 
≈ 0, 920
 15000
4300
.
15000
Op t = 0 is N = 15 000, dus b = 15 000.
Dus N = 15 000 ⋅ 0,920t.
15 a Rechte lijn op logaritmisch papier dus NI = b ⋅ gt.
1
 500  2
≈ 1, 291.
Lijn door (1, 300) en (3, 500), dus g = 
 300
t
NI = b ⋅ 1,291 .
300 = b ⋅ 1,291
voor t = 1 is NI = 300
300
=b
1, 291
b ≈ 230
}
Dus NI = 230 ⋅ 1,291t.
b Rechte lijn op logaritmisch papier dus NII = b ⋅ gt.
1
 400 2
Lijn door (1, 700) en (3, 400), dus g = 
≈ 0, 756.
 700 
NII = b ⋅ 0,756t
700 = b ⋅ 0,756
voor t = 1 is NII = 700
700
=b
0, 756
b ≈ 930
}
Dus NII = 930 ⋅ 0,756t.
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16 a gmaand = 1,002
1,002T = 2
T = 1,002 log( 2) =
log( 2)
≈ 347 maanden ≈ 29 jaar
log(1, 002)
Dus de verdubbelingstijd is 29 jaar.
b gweek = 0,8
0,
8T =
1
2
T = 0,8log
()
1
2
=
log
( )
≈
3,1 weken ≈ 22 dagen.
1
2
( )
log 0, 8
De halveringstijd is 22 dagen.
17
g32 dagen =
gdag =
()
1
2
1
2
1
32
≈ 0, 979
De afname is 2,1% per dag.
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