Bliggy · 2020. 9. 6. · Opgaven en indeling Opgaven 18.1, 18.2, 18.9, 18.10, 18.17, 18.18, extra....

Zomercursus Wiskunde BWeek 1, les 3

Jolien [email protected]

Korteweg-de Vries Instituut voor WiskundeFaculteit der Natuurwetenschappen, Wiskunde en Informatica

Universiteit van Amsterdam

6 juli 2017

Jolien Oomens Zomercursus Wiskunde B

Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.

Worteltrekken is het omgekeerde:

52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.

Regels voor machtsverheffen

ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24

= 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2

= 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b

= g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga

(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =

√25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒

2√

25 =

√25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b

⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Machten

Machtsverheffen:

24 = 2 · 2 · 2 · 2 = 16.

Regels zijn waar:

ga+b = g · g · · · · · g · g︸︷︷︸a+b keer

= g · · · · g︸︷︷︸a keer

· g · · · · g︸︷︷︸b keer

= gagb.


52 = 25 ⇒ 2√

25 =√

25 = 5

a3 = b ⇒ a =3√b.


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Exponentiele functies

Bekijk de functie

s

f (x) = 2x

, g(x) =(12

)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1

= 12

2−2

= 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1

= 11/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g



Bekijk de functie

s

f (x) = 2x

, g(x) =(12

)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2

= 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1

= 11/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g



Bekijk de functie

s

f (x) = 2x

, g(x) =(12

)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1

= 11/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g



Bekijk de functies f (x) = 2x , g(x) =(12

)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1

= 11/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2

(12

)2= 1

4

(12

)−1

= 11/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1

= 11/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1= 1

1/2

= 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1= 1

1/2 = 2

(12

)0= 1

(12

)−2

= 1(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1= 1

1/2 = 2

(12

)0= 1

(12

)−2= 1

(1/2)2

= 11/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1= 1

1/2 = 2

(12

)0= 1

(12

)−2= 1

(1/2)2= 1

1/4

= 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




)x.

x

y

-2 -1 1 2

-1

1

2

3

4

0

20 = 1

21 = 2

22 = 4

2−1 = 12

2−2 = 122

= 14

y = 2x

(12

)1= 1

2 (12

)2= 1

4

(12

)−1= 1

1/2 = 2

(12

)0= 1

(12

)−2= 1

(1/2)2= 1

1/4 = 4

y =(12

)x


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Exponentiele vergelijkingen

We willen oplossen 2x = a.

x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = − 1

2x = 3 ⇒ x = ? ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12

⇒ x = − 1

2x = 3 ⇒ x = ? ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = − 1

2x = 3 ⇒ x = ? ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = − 1

2x = 3

⇒ x = ? ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = − 1

2x = 3 ⇒ x = ?

≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = − 1

2x = 3 ⇒ x = ? ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = − 1

2x = 3 ⇒ x = 2log 3 ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g




x

y

-2 -1 1 2

1

2

3

4

0

y = 2x

a

a

?

We hebben:

2x = 12 ⇒ x = 2log 1

2 = −1

2x = 3 ⇒ x = 2log 3 ≈ 1.6


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes

De oplossing van de vergelijking 2x = a

isx = 2log a. Meer algemeen is x = g log a deoplossing van g x = a. Dus:

gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125

g log 1 = 0 want g0 = 1g log(−1) bestaat niet want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes

De oplossing van de vergelijking 2x = a isx = 2log a.

Meer algemeen is x = g log a deoplossing van g x = a. Dus:

gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes

De oplossing van de vergelijking 2x = a isx = 2log a. Meer algemeen is x = g log a

deoplossing van g x = a. Dus:

gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes

De oplossing van de vergelijking 2x = a isx = 2log a. Meer algemeen is x = g log a deoplossing van g x = a.

Dus:

gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes

De oplossing van de vergelijking 2x = a isx = 2log a. Meer algemeen is x = g log a deoplossing van g x = a. Dus:

gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9

= 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2

want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 9

5log 125 = −2 want 5−2 = 1

52= 1

25g log 1 = 0 want g0 = 1

g log(−1) bestaat niet want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25

= −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2

want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2

= 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125

g log 1

= 0 want g0 = 1g log(−1) bestaat niet want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125

g log 1 = 0

want g0 = 1g log(−1) bestaat niet want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125

g log 1 = 0 want g0 = 1

g log(−1) bestaat niet want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125

g log 1 = 0 want g0 = 1g log(−1)

bestaat niet want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125

g log 1 = 0 want g0 = 1g log(−1) bestaat niet

want g x > 0

1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2

= −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1

want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1

= 11/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2

= 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3

= 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12

want√

3 = 312


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Logaritmes


gg log a = a.

Voorbeelden:

3log 9 = 2 want 32 = 95log 1

25 = −2 want 5−2 = 152

= 125


1/2log 2 = −1 want(12

)−1= 1

1/2 = 2

3log√

3 = 12 want

√3 = 3

12


ga+b = gagb

g0 = 1

g−a = 1ga

gb−a = gb

ga(ga)b

= gab

(gh)a = gaha

g1/a = a√g


Grafiek van de logaritme

De logaritme y = 2log a is de oplossing van de vergelijking 2y = a.Bekijk de functie

s

f (x) = 2log x

, g(x) = 1/2log x

.

x

y

-1 1 2 3 4

-2

-1

1

2

0

2log 1 = 0

2log 2 = 1

2log 4 = 2

2log 12 = −1

2log 14 = −2

y = 2log x

y = 1/2log x


Grafiek van de logaritme

De logaritme y = 2log a is de oplossing van de vergelijking 2y = a.Bekijk de functies f (x) = 2log x , g(x) = 1/2log x .

x

y

-1 1 2 3 4

-2

-1

1

2

0

2log 1 = 0

2log 2 = 1

2log 4 = 2

2log 12 = −1

2log 14 = −2

y = 2log x

y = 1/2log x



Herinner

g x = a ⇒ x = g log a.

Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16

⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16

⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2

⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2

⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3

⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3

⇒ x2 = 12 ⇒ x = ±

√12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12

⇒ x = ±√

12 .



Herinner


Los op:

2x+3 = 16 ⇒ x + 3 = 2log 16 ⇒ x + 3 = 4

⇒ x = 1.

Zo ook

51−2x = 2 ⇒ 1− 2x = 5log 2 ⇒ 2x = 1− 5log 2

⇒ x = 12 −

12 ·

5log 2.

Ten slotte

9x2

= 3 ⇒ x2 = 9log 3 ⇒ x2 = 12 ⇒ x = ±

√12 .


Vergelijkingen met logaritmes

De vergelijking g log x = a heeft alsoplossing x = ga:

g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha

Regels voor logaritmes

g x = a ⇒ x = g log a

g log a + g log b = g log(ab)g log a− g log b = g log

(ab

)g log an = n g log a

g log a =hlog ahlog g

.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha



g log a + g log b = g log(ab)

g log a− g log b = g log(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab

)

g log an = n g log a


.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.



De vergelijking g log x = a

heeft alsoplossing x = ga:

g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga

= a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3

⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2

= 10. Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10.

Verder:

2log(2x − 2) = 2− 2log x2log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) = 2− 2log x

2log(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) + 2log x = 2

2log(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

2log(x − 2) = 3 ⇒ x − 2 = 23,

dus x = 23 + 2 = 10. Verder:

2log(2x − 2) + 2log x = 22log

(x(2x − 2)

)= 2

2log(2x2 − 2x) = 2

2x2 − 2x = 22

x2 − x − 2 = 0.


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

3log(x − 1) + 1/3log x = 1

3log(x − 1) +3log x

3log 1/3= 1

3log(x − 1) +3log x−1 = 1

3log(x − 1)− 3log x = 13log x−1

x = 1x−1x = 3

x − 1 = 3x


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

3log(x − 1) + 1/3log x = 1


3log 1/3= 1

3log(x − 1) +3log x−1 = 1

3log(x − 1)− 3log x = 1

3log x−1x = 1

x−1x = 3

x − 1 = 3x


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

3log(x − 1) + 1/3log x = 1


3log 1/3= 1

3log(x − 1) +3log x−1 = 1

3log(x − 1)− 3log x = 13log x−1

x = 1

x−1x = 3

x − 1 = 3x


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.




g log ga = a.

Voorbeelden:

3log(x − 1) + 1/3log x = 1


3log 1/3= 1

3log(x − 1) +3log x−1 = 1

3log(x − 1)− 3log x = 13log x−1

x = 1x−1x = 3

x − 1 = 3x


ga+b = gagb

g−a = 1ga(

ga)b

= gab

(gh)a = gaha




(ab



.


Opgaven en indeling

Opgaven

18.1, 18.2, 18.9, 18.10, 18.17, 18.18, extra.

Antwoorden van de opgaven staan achterin, uitwerkingen van deextra opgaven op http://www.bliggy.net/cursusB.html.

Groepen

De indeling is op basis van je achternaam:

A t/m D: zaal A1.08 (Gideon Jager)

E t/m Kuhl: zaal D1.115 (Jeroen Eijkens)

Kuhlhan t/m Seydel: zaal D1.113 (Sebastian Zur)

Simsir t/m Z: zaal D1.112 (Thijs Benjamins)


Bliggy · 2020. 9. 6. · Opgaven en indeling Opgaven 18.1, 18.2, 18.9, 18.10, 18.17, 18.18, extra....

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Transcript of Bliggy · 2020. 9. 6. · Opgaven en indeling Opgaven 18.1, 18.2, 18.9, 18.10, 18.17, 18.18, extra....