Chapter 8

Logarithmic and Exponential Functions

This chapter contains logarithmic and exponential functions as a continuous study of those of logarithms and exponents of numbers from grade 10.

8.1 Logarithmic Functions

For a number b > 0, b ≠ 1, the logarithmic function y = logb x is defined as y is a number such that x = by for x > 0.
The graphs of y = log2 x are as follows.
fig 8.1-1

fig 8.1-2

Domain of y = logbx: x > 0
Range: ℝ
Asymptote: y-axis (x = 0)
For b > 1, y = logbx is an increasing function that means if x1 > x2 then y1 = logbx1 > y2 = logbx2. For example, y = log2x is an increasing function.
For 0 < b < 1, y = logbx is a decreasing function that means if x1 > x2 then y1 = logb x1 < y2 = logbx2. For example, y = log 1 ⁄2 x is a decreasing function.
Since log 1 ⁄b x = - log 1 ⁄b x,
$$y = \text{log}_b\, x \xrightarrow[\text{on} \, x\text{-axis}] {\text{reflection}} \, y = \text{log}_{\tiny{\dfrac{1}{2}}}\, x $$ as the graphs of y = log2x and y = log 1 ⁄2 x.

For b > 1, the graph of y = logbx is vertical scaling with scale factor 1 ⁄log2b of the graph y = log2x.

logb x = log2 x⁄log2 b so that $$y = \text{log}_2\, x \xrightarrow[\text{scale factor} \, \tiny{\dfrac{1}{\text{log}_2 b}}] {\text{reflection}} \, y = \text{log}_b\, x $$
For 0 < b < 1, the graph of y = logb x is vertical scaling with scale factor 1⁄log 1⁄2 b of the graph y = 1⁄log 1⁄2 x .
logb x = 1⁄log 1⁄2 x ⁄ 1⁄log 1⁄2 b so that
$$y = \text{log}_{\tiny{\dfrac{1}{2}}}\, x \xrightarrow[\text{scale factor} \, \tiny{\dfrac{1}{\text{log}_{\tiny{\dfrac{1}{2}}} b}}] {\text{reflection}} \, y = \text{log}_b\, x $$ Therefore all graphs of y = logb x are vertical scaling of the graph of y = log2 x or th graph of y = log 1⁄2 x. So as in y = log2 x and y = log 1⁄2 x, we have
b > 1 0 < b < 1
y = logb x < 0,     if 0 < x < 1
y = logb x = 0,     if x = 1
y = logb x > 0,     if x > 1 		y = logb x > 0,     if 0 < x < 1
y = logb x = 0,     if x = 1
y = logb x < 0,     if x > 1


Example 1.
From the graph of y = log2 x, draw step-by-step transformation graph to get the graph of (a) y = log2 (x - 1) + 2
(b) y = log1⁄2 (x + 2) - 1.
Solution
$$\text{(a)}\, \, \, y = \text{log}_2 x \xrightarrow[\text{vertical translation 2 units}] {\text{horizontal translation 1 unit}} \, y = \text{log}_2\, (x - 1) + 2 $$
fig 7.1-3

$$\text{(b)} \, \, \, y = \text{log}_2\, x \xrightarrow[\text{on}\, x\text{-axis}] {\text{reflection}} \, y $$ $$ = \text{log}_{\tiny\dfrac{1}{2}}\, x \, \xrightarrow[\text{vertical translation -1 unit}] {\text{horizontal translation -2 units}} \, y $$ $$ = \text{log}_{\tiny{\dfrac{1}{2}}} (x + 2) -1$$
fig 8.1-4

Example 2
Draw the graph of y = 2 log2 x     and     y = log2 x2
Solution
The graph of y = 2 log2 x can be drawn directly or can be drawn as the vertical scaling with scale factor 2 of the graph of y = log2 x. Domain of y = log2 x2 is ℝ \ {0} and y = log2 x2 is an even function.
fig 8.1-5

Note: Asymptote y-axis (x = 0)
fig 8.1-6


Note: Asymptote y-axis (x = 0)
Note that 2 log2 x = log2 x2 for x > 0.

Exercise 8.1
1. Draw the graph of
(a)   y = log2(x - 2) + 1
Solution
$$ y = \text{log}_2 x \xrightarrow[\text{vertical translation 1 unit}] {\text{horizontal translation 2 units}} \, y = \text{log}_2\, (x - 2) + 1 $$
fig 8.1-7

Domain = {x | x > 2}, Range = R

(b)  y = log2 (x + 1) - 2
Solution
$$ y = \text{log}_2 x \xrightarrow[\text{vertical translation -2 units}] {\text{horizontal translation -1 unit}} \, y = \text{log}_2\, (x + 1) - 2 $$
fig 8.1-8

Domain = {x| x > -1}, Range = R

(c)  y = log1⁄ 2 (x + 1) + 2
Solution
$$ y = \text{log}_2 x \xrightarrow[\text{on} \, x-\text{axis}] {\text{reflection}} \, y $$ $$ = \text{log}_{\frac{1}{2}} x \xrightarrow[\text{vertical translation 2 units}] {\text{horizontal translation -1 unit}} \, y = \text{log}_{\frac{1}{2}}\, (x + 1) + 2 $$
fig 8.1-9

Domain = {x | x > -1}, Range = R

(d)  y = log1⁄ 2 (x - 2) - 1
Solution
$$ y = \text{log}_2 x \xrightarrow[\text{on} \, x-\text{axis}] {\text{reflection}} \, y $$ $$ = \text{log}_{\frac{1}{2}} x \xrightarrow[\text{vertical translation -1 units}] {\text{horizontal translation 2 unit}} \, y = \text{log}_{\frac{1}{2}}\, (x - 2) - 1 $$
fig 8.1-10

Domain = {x | x > 2}, Range = R

2. Draw the graph of
(a)   y = log1⁄ 2 x2
Solution
fig 8.1-12

asymptote x = 0
Domain = R, Range = R

(b) y = log1⁄ 2 x
Solution
fig 8.1-13

asymptote x = 0,
Domain = {x| x > 0}
Range = R

3. Draw the graph of
(a)   y = log2(-x)
Solution
fig 8.1-14

asymptote x = 0,
Domain = {x | x < 0}
Range = R

(b)   y = log1⁄ 2 (-x)

Solution
fig 8.1-15

asymptote x = 0,
Domain = {x | x < 0}
Range = R

4. Draw the graph of
(a)   y = log2 |x|
Solution
fig 8.1-16

asymptote x = 0,
Domain = R
Range = R

(b)   y = log1⁄ 2 |x|
Solution
fig 8.1-17

asymptote x = 0,
Domain = R
Range = R