Section 11-4
Logarithmic Functions
Objective: Students will be able to
1. Evaluate expressions involving logarithms
2. Solve equations involving logarithms
3. Graph logarithmic functions and inequalities
Since the graphs of exponential functions pass the horizontal line test, we know their
inverses are also functions.
Remember to find the inverse of a graph, we can switch the x & y variables around and
then graph them.
The graphs are reflections of
y = 2𝑥
Inverse
each other across the line y = x.
x y
x y
0 1
1 0
1 2
2 1
1
1
-1 2
-1
2
The inverse of an exponential is called a logarithm, and is written as log 𝑏 𝑦 = 𝑥.
Example 1:
a.
Write each equation in exponential form.
3
log32 8 = 5
3
5
32 = 8
b.
1
log 81 3 = 4
1
4
81 = 3
Example 2: Write each equation in logarithmic form.
𝑎.
64 = 1296
𝑏.
1
log 6 1296 = 4
Example 3:
a.
1
2−8 = 256
log 2 256 = −8
Evaluate each expression.
1
log 3 27
1
log 3 27 = x
b.
log4 45
log 4 45 = x
4𝑥 = 45
1
3𝑥 = 27
x=5
3𝑥 = 3−3
x = -3
c.
log4 8
log 4 8 = 𝑥
4𝑥 = 8
22
𝑥
= 23
2𝑥 = 3
3
𝑥=
2
d.
1
log16 4
1
log16 4 = x
1
16𝑥 = 4
24
𝑥
= 2
4𝑥 = −2
1
𝑥=−
2
−2
Properties of Logs
Sum of Logs
Difference of Logs
Constant times a Log
Equality
log 𝒃 𝒎𝒏 = log 𝒃 𝒎 + log 𝒃 𝒏
𝑚
log 𝑏 = log 𝑏 𝑚 − log 𝑏 𝑛
𝑛
log 𝑏 𝑚𝑝 = 𝑝 log 𝑏 𝑚
log 2 12 = log 2 4 + log 2 3
1
log 3 9 = log 3 1 − log 3 9
log 2 8𝑥 = 𝑥 log 2 8
𝐼𝑓 log 𝑏 𝑚 = log 𝑏 𝑛 , 𝑡ℎ𝑒𝑛 𝑚 = 𝑛 log 8 3𝑥 − 4 = log 8 5𝑥 + 2 ,
then 3x - 4 = 5x + 2
To Solve Log Equations:
Type 1: Logs on both sides
1.
Simplify on both sides to get one log
on each side. (Use Prop. Of Logs.)
Type 2: Logs on one side
1.
Simplify to get only one log in the
problem.
2.
Use Prop. Of Equality and drop the logs.
2. Rewrite problem as an exponent.
3.
Solve for the variable.
3. Solve for the variable.
4.
Check your answers.
4. Check your answers.
Example 4: Solve each equation.
𝑎.
1
4
log 𝑝 6561 =
1
2
=
𝑝=
𝑝=
log 5 5𝑥 − 3 = log 5 10𝑥 + 2
5x – 3 = 10x + 2
1 2
65614
-3 = 5x + 2
-5 = 5x
1
65612
6561
𝑝 = ± 81
c.
b.
1
4
𝑝 = 6561
1 2
𝑝2
1
2
-1 = x
log8 (x + 1) + log8 (x + 3) = log8 24
log 8 𝑥 + 1 𝑥 + 3 = log 8 24
𝑥 2 + 4x + 3 = 24
𝑥 2 + 4x – 21 = 0
(x + 7)(x – 3) = 0
x = -7, 3
3
∅
P = 81
d.
log3 3x = -1
3−1 = 3x
1
3
= 3x
1
9
=x
e.
log10 4 + 2 log10 x = 2
f.
log10 4𝑥 2 = 2
102 = 4𝑥 2
log3 (x + 3) – log3 (2x – 1) = log3 2
𝑥+3
log 3 2𝑥−1 = log 3 2
𝑥+3
=
2𝑥−1
100 = 4𝑥 2
2
𝑥 + 3 = 2(2𝑥 − 1)
25 = 𝑥 2
𝑥 + 3 = 4𝑥 − 2
±5 = 𝑥
To Graph Log Functions….
5 = 3𝑥
5
3
1)
Rewrite the logarithm as an exponential
2)
Make an xy-table of values for the parent function. Pick numbers for y and solve
for x.
3)
Shift the parent function based on a, b, h, and k.
4)
State the V.A., Domain and Range.
𝑦 = 𝑎 log 𝑐 𝑏 𝑥 − ℎ + 𝑘
=x
Example 5:
Graph y = log4 (x + 2).
𝑃𝑎𝑟𝑒𝑛𝑡 𝐺𝑟𝑎𝑝ℎ:
4𝑦 = x
h = -2
𝑦 = log 4 𝑥
x y
1 0
4 1
1
-1
4
x
-1
2
7
-4
y
0
1
−1
D: x > -2
R: R
V.A. : x = -2
Example 6:
Graph y = log3 x - 4.
𝑃𝑎𝑟𝑒𝑛𝑡 𝐺𝑟𝑎𝑝ℎ:
3𝑦 = x
𝑦 = log 3 𝑥
x y
1 0
3 1
1
-1
3
x y
1 -4
3 -3
1
-5
3
k = -4
D: x>0
R: R
V.A. : x = 0
Example 7:
Graph 𝑦 =
1
2
ln 𝑥 + 1 - 3
1
𝑃𝑎𝑟𝑒𝑛𝑡 𝐺𝑟𝑎𝑝ℎ: 𝑦 = ln 𝑥
𝑒𝑦= x
𝑥 𝑦
1
0
2.718 1
0.368 − 1
a=2
h = -1
k = -3
𝑥
1
𝑦
0
1
2.718
2
1
0.368 −
2
𝑥
0
𝑦
−3
5
1.718 −
2
7
−0.632 −
2
D: x > -1
R: R
V.A.: x = -1
Example 8: Graph the following logarithmic inequality.c inequality
y ≥ log 3 (𝑥 + 1)
Parent Graph:
𝑦 = log 3 𝑥
3𝑦 = x
𝑥
1
9
1
3
1
3
9
𝑦
-2
-1
0
1
2
𝑥
8
-9
Horizontal translation 1 lt.
𝑦
-2
2
- 3 -1
0
0
2
1
8 2
Test (1, 2)
2 ≥ log 3 (1 + 1)
2 ≥ log 3 2
2 ≥ 1.58496
𝑇𝑟𝑢𝑒