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Logarithmic Functions

Logarithmic Functions. f(x)= alog c b(x-h) + k. If c > 1. a > 0. a < 0. b > 0. b < 0. If 0 < c < 1. a > 0. a < 0. b > 0. b < 0. Logarithmic Functions. The rule f(x)= alog c b(x-h) + k can be re-written in the form: f(x)= log c b(x-h)

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Logarithmic Functions

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  1. Logarithmic Functions f(x)= alogcb(x-h) + k

  2. If c > 1 a > 0 a < 0 b > 0 b < 0

  3. If 0 < c < 1 a > 0 a < 0 b > 0 b < 0

  4. Logarithmic Functions The rule f(x)= alogcb(x-h) + k can be re-written in the form: f(x)= logcb(x-h) where the function passes through the point and has x = h as its asymptote x-intercept

  5. Graphing Logarithmic Functions

  6. Example 1 y = 2log3(x-1) + 2 Step 1 – Draw the vertical asymptote x = h x = 1

  7. y = 2log3(x-1) + 2 Step 2 – Locate two points using the rule Let x=2 y = 2(0) + 2 y = 2 A(2,2)

  8. y = 2log3(x-1) + 2 Step 2 – Locate two points using the rule Let x=4 y = 2(1) + 2 y = 4 A(4,4)

  9. y = 2log3(x-1) + 2Step 3 – Sketch the curve

  10. Example 2 y = 3log0.5(x+4) - 3

  11. Example 3 y = -log1/3(-0.5x+2) + 1 y = -log1/3-0.5(x-4) + 1

  12. Finding the rule of Logarithmic Functions

  13. x intercept,1 point and an asymptote f(x)=logcb(x-h) Step 1: Find the x intercept and determine parameter b using Step 2: Substitute the point (x,y) and asymptote (h) Step 3: Solve for “c”

  14. f(x)=logcb(x-h) Example 1 1 +h = 2.5 b x = 0.5 y = 2 1 +0.5 = 2.5 b f(x)=log0.50.5(x-0.5) (0,4) 1 = 2 b (4,-2) b = 0.5 f(x)=logcb(x-h) 2 = logc0.5(1-0.5) 2 = logc(0.25) c2 = 0.25 c = +/- 0.5

  15. f(x)=logcb(x-h) Example 2 1 +h = -2 b f(x)=log525(x+2) y = 2 1 -2 = -1.96 b 1 = 0.04 b b = 25 f(x)=logcb(x-h) 3.23 = logc25(5.24+2) 3.23 = logc(181) c3.23 = 181 c = +/- 5

  16. Inverse of a Logarithm Example 3:Find the inverse of y = 3log24(x-1) - 6 x =3log24(y-1) - 6 x+6 =3log24(y-1) x+6 =log24(y-1) 3

  17. Inverse of a Logarithm Example 3:Find the inverse of y = 5ln3(x-1) - 10 x =5ln3(y-1) - 10 x+10 =5ln3(y-1) x+10 =ln3(y-1) 5

  18. Inverse of an Exponential Example 4:Find the inverse of

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