Powerpoint Templates Page 1

Logarithmic Functions

f(x)= alogcb(x-h) + k

If c > 1

b > 0

a < 0a > 0

b < 0

If 0 < c < 1

b > 0

a < 0a > 0

b < 0

Powerpoint Templates Page 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

Powerpoint Templates Page 5

Graphing Logarithmic Graphing Logarithmic FunctionsFunctions

Example 1 y = 2log3(x-1) + 2

Step 1 – Draw the vertical asymptote x = h

x = 1

Step 2 – Locate two points using the rule

Let x=2

y = 2(0) + 2

y = 2

A(2,2)

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)

y = 2log3(x-1) + 2

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

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

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

y = -log1/3-0.5(x-4) + 1

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Finding the rule of Finding the rule of Logarithmic FunctionsLogarithmic Functions

x intercept,1 point and an asymptote

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”

f(x)=logcb(x-h)

Example 1

(0,4)

(4,-2)

f(x)=logcb(x-h)1 +h = 2.5b

b = 0.5

f(x)=logcb(x-h)

2 = logc0.5(1-0.5)

2 = logc(0.25)

y = 2

c2 = 0.25

c = +/- 0.5

1 +0.5 = 2.5b1 = 2b

f(x)=log0.50.5(x-0.5)

x = 0.5

Example 2 f(x)=logcb(x-h)1 +h = -2b

b = 25

f(x)=logcb(x-h)

3.23 = logc25(5.24+2) 3.23 = logc(181)

y = 2

c3.23 = 181

c = +/- 5

1 -2 = -1.96b1 = 0.04b

f(x)=log525(x+2)

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Inverse of a Logarithm

x =3log24(y-1) - 6

Example 3: Find the inverse of y = 3log24(x-1) - 6

x+6 =3log24(y-1)x+6 =log24(y-1)

3

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Inverse of a Logarithm

x =5ln3(y-1) - 10

Example 3: Find the inverse of y = 5ln3(x-1) - 10

x+10 =5ln3(y-1)x+10 =ln3(y-1)

5

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Inverse of an ExponentialExample 4: Find the inverse of