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 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) + 2 Step 3 – Sketch the curve

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

11. Example 3 y = -log1/3(-0.5x+2) + 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. Example 1 f(x)=logcb(x-h) f(x)=log0.50.5(x-0.5) 1 +h = 2.5 b x = 0.5
y = 2
= 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. Example 2 f(x)=logcb(x-h) f(x)=log525(x+2) 1 +h = -2 b 1 -2 = -1.96 b
y = 2
1 -2 = b
1 = 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 x =3log24(y-1) - 6 x+6 =3log24(y-1)
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)

17. Inverse of a Logarithm x =5ln3(y-1) - 10 x+10 =5ln3(y-1)
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)

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