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Samenvatting Wiskunde: logaritmische functies

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Samenvatting Wiskunde Logaritmische functies 5de middelbaar (Lyceum) Samenvatting is ordelijk en volledig Bevat theorie en voorbeeldoefeningen met opgave, uitwerking en stappenplan

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  • January 21, 2023
  • 3
  • 2020/2021
  • Summary
  • Secondary school
  • 3e graad
  • 5
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Logaritmische functies

● Definitie
⇒ alog x= y ⇔ ay= x met a ∈ R+0 \ {1} en x ∈ R+0
uj

⇒ a= grondtal
⇒ • a moet strikt positief zijn en mag niet 0 of 1 zijn
⇒ We kunnen alleen logaritmen van strikt positieve getallen berekenen.

● Rekenregels
a
1) Logaritme van een product: log (x . y)= alog x + alog y
a 𝑥
2) Logaritme van een quotiënt: log 𝑦 = alog x - alog y
a
3) Logaritme van een macht: log xn= n . alog x
𝑏
a 𝑙𝑜𝑔 𝑥
4) Verandering van grondtal: log x= 𝑏
𝑙𝑜𝑔 𝑦


● Gevolgen
a
⇒ log ay= y want vervang in alog x= y gewoon x door ay (ay = x)
𝑎
⇒ a 𝑙𝑜𝑔 𝑥= x want vervang in ay= x gewoon y door alog x (alog x = y)
a
⇒ log a= 1 want a1= a
a
⇒ log 1= 0 want a0= 1

● De 10-delige of gewone logaritme
⇒ De logaritme met grondtal a = 10
⇒ Als er geen grondtal vermeld wordt is het grondtal 10
⇒ Voorbeeld: log 100= 10log 10²= 2

● Voorbeeld oefening (Oefening 1b, 1c, 1f, 1g, 1i e n 1j pagina 152)
⇒ reken uit
b) ²log 16= ²log 24= 4 . ²log 2= 4 → RR3
1
1 1
c) ³log 3= ³log 3 = 2
2
. ³log 3= 2
→ RR3
f) 5log (-25)= / → -25<0
g) log10 2= 10log 10 2= 2. 10log 10= 2 → RR3
1
i) ²log 4
= ²log 1 - ²log 4= 0 - ²log 2²= 0 - 2= -2 → RR2
j) 25log 1= 0 → alog 1= 0

● Voorbeeld oefening (Oefening 2a, 2c en 2d pagina 152)
⇒ Los op met behulp van logaritmen
a ) 2x= 32 ⇔ x= ²log 32= ²log 25= 5 . ²log 2= 5
1
1 1 1 1 1 1
c) ( 2 )x= 4 ⇔ x= 2 log 4= 2 log 2²= 2 log ( 2 )-2= -2 . 2 log 2
= -2 . 1= -2
1 1 1
d) 7x= 49
⇔ x= 7log ( 49 )= 7log 7²
= 7log 7-2= -2 . 7log 7= -2

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