# havo B 9.3 Logaritmische functies - PowerPoint PPT Presentation

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havo B 9.3 Logaritmische functies. Logaritme en exponent. 2 x = 8 x = 3 want 2 3 = 8 2 x = 8 ⇔ 2 log(8) 2 3 = 8 ⇔ 2 log(8) = 3 2 log(32) = 5 want 2 5 = 32 algemeen: g log( x ) = y betekent g y = x dus g log( g y ) = y x > 0 , g > 0 en g ≠ 0.

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havo B 9.3 Logaritmische functies

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havo B 9.3 Logaritmische functies

### Logaritme en exponent

• 2x = 8

• x = 3 want 23 = 8

• 2x = 8 ⇔ 2log(8)

• 23 = 8 ⇔2log(8) = 3

• 2log(32) = 5 want 25 = 32

• algemeen:

• glog(x) = y betekent gy = x

• dus glog(gy) = y

• x > 0 , g > 0 en g ≠ 0

a 5log(0,2) =

5log() =

5log(5-1) =

-1

b3log(3√3) =

3log(31 . 3½) =

3log(31½) =

c½log(8) =

½log((½)-3) =

-3

d¼log() =

¼log((¼)2) =

2

### De standaardgrafiek y = glog(x)

functies f en g met de eigenschap dat hun grafieken elkaars

spiegelbeeld zijn in de lijn y = x heten inverse functies

g > 1

0 < g < 1

y

y

y = x

y = x

y = 2x

1

y = (½)x

1

x

x

O

O

1

1

y = 2log(x)

y = ½log(x)

### voorbeeld

x = 4

y

4

ay = 3log(x)

4 naar rechts

y = 3log(x – 4)

2 omhoog

y = 3log(x – 4) + 2

b

Df= < 4, >

3

2

1

x

1

3

9

3log(x)

-1

0

1

2

-2

O

5

1

2

3

4

2 omhoog

-1

4 naar rechts

-2

### opgave 44

4

averticale asymptoot :

4x – 1 = 0

x = ¼

voer in y1 = log(4x-1)/log(3)

bf(x) ≤ 2

3log(4x – 1) = 2

4x – 1 = 32

4x = 10

x = 2½

¼ < x ≤ 2½

3

y = 2

2

x

1

2

3

4

1

1,8

2,2

2,5

1

3log(4x - 1)

-1

0

2

3

4

x

1

-1

-2

x = ¼

### opgave 47

af(x) = 6 + ½log(x2 + 5)

x2 + 5 = 0 heeft geen oplossingen

dus f heeft geen verticale asymptoot

g(x) = 3log(x2 – 2x)

x2 – 2x = 0

x(x – 2) = 0

x = 0 v x = 2

voer in y1 = 6 + log(x2 + 5)/log(½)

en y2 = log(x2 – 2x)/log(3)

y

g

f

x

O

x = 0

x = 2

boptie intersect

(-2,759 ; 2,344) en (3,776 ; 1,732)

cf(x) > g(x)

-2,759 < x < 0 v 2 < x < 3,776

g

f

x

O

-2,759

3,776

x = 0

x = 2
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