Logarithmic Functions f(x)= alogcb(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)= 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

Graphing Logarithmic Functions

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

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)

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 Step 3 – Sketch the curve

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

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

Finding the rule of Logarithmic Functions

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”

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 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

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 = -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

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) 3

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) 5

Inverse of an Exponential Example 4: Find the inverse of