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havo B 9.3 Logaritmische functies

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.

richard-gibbs
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havo B 9.3 Logaritmische functies

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  1. havo B 9.3 Logaritmische functies

  2. 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

  3. a 5log(0,2) = 5log() = 5log(5-1) = -1 b3log(3√3) = 3log(31 . 3½) = 3log(31½) = 1½ c½log(8) = ½log((½)-3) = -3 d¼log() = ¼log((¼)2) = 2 voorbeeld

  4. 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)

  5. voorbeeld x = 4 y 4 a y = 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  

  6. opgave 44 4 a verticale 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 2½ -1 -2 x = ¼

  7. opgave 47 a f(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

  8. b optie intersect (-2,759 ; 2,344) en (3,776 ; 1,732) c f(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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