,Hoofdstuk 9 Exponentiële en Logaritmische functies (V5 Wis B)Pagina 2 van 19
VOORBEELD 2
Bereken exact
5
a. log(2𝑥 − 1) = 3
b. 4 ∙ 2log(2𝑥) + 1 = 13
OPLOSSING 2
Je kunt dit op twee manieren oplossen :
𝑔
Met de hoofdregel : 𝑔𝑡 = 𝑏 ⇔ 𝑡 = log(𝑏)
𝑔
Met de hulpregel : log(𝑔𝑡 ) = 𝑡
5 5
a. log(2𝑥 − 1) = 3 Of log(2𝑥 − 1) = 3
3 5
2𝑥 − 1 = 5 Of log(2𝑥 − 1) = 3
5
2𝑥 − 1 = 125 Of log(2𝑥 − 1) = 5log(53 )
2𝑥 = 126 Of 2𝑥 − 1 = 125
𝑥 = 63 Of 2𝑥 = 126
𝑥 = 63
b. Deze doen we alleen met de hoofdregel
4 ∙ 2log(2𝑥) + 1 = 13
4 ∙ 2log(2𝑥) = 12
2
log(2𝑥) = 3
2𝑥 = 23
2𝑥 = 8
𝑥=4
, Hoofdstuk 9 Exponentiële en Logaritmische functies (V5 Wis B)Pagina 3 van 19
LES 2 : GRAFIEKEN EN ONGELIJKHEDEN
STAPPENPLAN LOGARITMISCHE ONGELIJKHEDEN
(0) Bepaal het domein
(1) Los de vergelijking op (I)
(2) Maak een schets van de twee grafieken. Let op het domein!!!! (S)
(3) Lees de oplossing af uit de schets. (A)
VOORBEELD 1
4 4
Gegeven 𝑓(𝑥) = 𝑙𝑜𝑔 (𝑥 + 2) en 𝑔(𝑥) = 𝑙𝑜𝑔 (5 − 𝑥).
a. Bepaal de domeinen van 𝑓 en 𝑔.
b. Los op : 𝑓(𝑥) ≤ 𝑔(𝑥)
OPLOSSING 1
a. 𝐷𝑜𝑚𝑒𝑖𝑛 𝑓 ∶ 𝑥 +2 >0
𝑥 > −2 (𝑉𝐴 ∶ 𝑥 = −2)
𝐷𝑜𝑚𝑒𝑖𝑛 𝑔 ∶ 5– 𝑥 > 0
𝑥 < 5 (𝑉𝐴 ∶ 𝑥 = 5)
4
b. (1) 𝑙𝑜𝑔 (𝑥 + 2) = 4𝑙𝑜𝑔 (5 − 𝑥).
𝑥+2=5−𝑥
2𝑥 = 3
1
𝑥 = 12 { of met de GR en intersect als er geen exact/alg. staat}
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