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Chapter 12
Logarithmic and
Exponential
Functions
§12.2
The Logarithmic
Function
Logarithms
The logarithm, base b, of a positive number x is the power
(exponent) to which the base b must be raised to produce x.
That is, y = logbx is the same as x = by, where b > 0 and b  0.
Exponent = y
x = by
y = logbx
Base = b
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #3
Exponential Equations in Log Form
Example:
Write the following exponential equations in logarithmic form.
Exponential Form
23 = 8
Logarithmic Form
log28= 3
6 2 = 1
36
log6 1 = 2
36
50 = 1
log51= 0

1
2
4
= 1
16
log1 2 1 = 4
16
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #4
Log Equations in Exponential Form
Example:
Write the following logarithmic equations in exponential form.
Logarithmic Form
log232= 5
Exponential Form
25 = 32
log4 1 = 3
64
4 3 = 1
64
log1010= 1
101 = 10
log1 3 1 = 2
9
  = 91
1
3
2
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #5
Solving Logarithmic Equations
Example:
Solve the following logarithmic equation.
a) logx 81 = 4
a) logx 81 = 4
x4 = 81
x4 = 34
x=3
b) log10 x = 2
Convert into an equivalent exponential equation.
Solve the exponential equation.
b) log10 x = 2
10– 2 = x
1
x
100
Convert into an equivalent exponential equation.
Simplify.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #6
Graphing a Logarithmic Function
Example:
Graph y = log2 x.
y = 2x
y
The exponential form of
the function is x = 2y.
x
y
1
4
1
2
2
1
0
2
1
4
2
1
2
y = log2 x
x
–2
2
2
y = log2 x is the inverse function of y = 2x.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #7
§12.3
Properties of
Logarithms
The Logarithm of a Product
Property 1: The Logarithm of a Product
For any positive real numbers M and N and any positive base b  1,
logbMN = logbM + logbN.
Example: Write the following logarithm as a sum of logarithms
a) log5(4 · 7)
b) log10(100 · 1000)
a) log5(4 · 7) = log54 + log57
b) log10(100 · 1000) = log10100 + log101000 = 2 + 3 = 5
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #9
The Logarithm of a Product
Example: Write the following logarithms as single logarithms
a) log24 + log2x
b) log 50 + log x + log 2
a) log24 + log2x = log24x
b) log 50 + log x + log 2 = log 50(x)(2) = log 100x
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #10
The Logarithm of a Quotient
Property 2: The Logarithm of a Quotient
For any positive real numbers M and N and any positive base b  1,
M
logb
 logb M  logb N.
N
 
Example: Write the following as a difference of logarithms.
45
z 2
a) log2
b) log z
11
3


z 2
=log (z  2)  log 3
log 

3
a) log2
b)
 
z
45
=log2 45 log2 11
11
z
z
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #11
The Logarithm of a Quotient
Example: Write the following as a single logarithm.
a) log5 12  log5 2
a) log5 12  log5 2 =log5
b) log c w2  log c y

12
=log5 6
2
2


w
2
b) log c w  log c y =log c  y 
 
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #12
The Logarithm of a Number Raised to a Power
Property 3: The Logarithm of a Product
For any positive real numbers M, and any real number p, and any
positive base b  1,
logbMp = plogbM.
Example: Write the following as a single logarithm.
a) 3 log8 5  log8 z
b) 2 log 100 + 3 log 10
z
a) 3 log8 5  log8 z = log8 53  log8 z = log8 53  log8 z =log8 125
b) 2 log 100 + 3 log 10 = log 1002 + log 103  log[10000 (1000)]
 log10000000  7
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #13
Solving Logarithm Equations
The following properties are true for all positive values of b  1, and
all positive values of x and y.
Property 4
logb b = 1
Property 5
logb 1 = 0
Property 6
If logb x = logb y, then x = y.
Example: a) Evaluate log9 9.
b) Evaluate log8 1.
c) Find y if log4 y = log4 19.
a) log9 9 = 1
b) log8 1 = 0
c) If log4 y = log4 19, then y = 19.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #14
Solving Logarithm Equations
Example: Solve for x.
log5 1 = log5 x – log5 8
log5 1 = log5 x – log5 8
log5 1 + log5 8 = log5 x
0 + log5 8 = log5 x
8=x
Isolate the variable.
Property 5
Property 6
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #15
Solving Logarithm Equations
Example: Solve for x.
log3(4x + 6) – log3(x – 1) = 2
log3(4x + 6) – log3 (x – 1) = 2
4x  6
log3
2
x 1
4x  6
32 
x 1
4x  6
9
x 1
9( x 1)  4x  6
9x  9  4x  6
5x 15
x 3
Property 1
Convert to exponential form.
Simplify.
Solve for x.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #16
§12.5
Exponential and
Logarithmic Equations
Solving Logarithmic Equations
When solving logarithmic equations, we generally try to get all of
the logarithms on one side of the equation and the numerical values
on the other side. Then the properties of logarithms are used to
obtain a single logarithmic expression on one side.
Step 1: If an equation contains some logarithms and some
terms without logarithms, try to get one logarithm
alone on one side and one numerical value on the
other.
Step 2: Convert to an exponential equation using the
definition of a logarithm.
Step 3: Solve the equation.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #18
Solving Logarithmic Equations
Example:
Solve log24 + log2(x – 1) = 5.
log24 + log2(x – 1) = 5
log2[4(x – 1)] = 5
log2(4x – 4) = 5
Property 1
Simplify.
4x – 4 = 25
Write in exponential form.
4x – 4 = 32
Solve for x.
4x = 36
x=9
Check:
log24 + log2(9 – 1) = 5
log24 + log28 = 5
2+3=5 
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #19
Solving Logarithmic Equations
Example:
Solve log(x + 3) + log x = log 4.
log(x + 3) + log x = log 4
log(x + 3)x = log 4
Property 1
(x + 3)x = 4
Property 6
x2 + 3x = 4
Simplify.
x2 + 3x – 4 = 0
(x + 4)(x – 1) = 0
x = – 4 or x = 1
Solve for x.
Stop! It is not possible
to take the logarithm of
a negative number.
Check:
log (4 + 3) + log(4) = log 4
log (1 + 3) + log 1 = log 4
log 4 + 0 = log 4 
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #20
Solving Exponential Equations
Property 7
If x and y > 0 and x = y, then logbx = logby, where b > 0 and b  1.
Notice that this is the reverse of Property 6.
Property 7 is also referred to as “taking the logarithm of each side
of the equation.”
Example:
Solve 5x = 23.
5x = 23
log 5x = log 23
Property 7
x log 5 = log 23
x
log 23
 1.948
log5
Property 3
Divide both sides by log 5.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #21
Solving Exponential Equations
Example:
Solve 153x – 2 = 230.
153x – 2 = 230
log 153x – 2 = log 230
(3x – 2)log 15 = log 230
log 230
3x  2 
log15
3x – 2 = 2.008116
3x = 4.008116
x  1.336
Property 7
Property 3
Divide each side by log 15.
Simplify.
Solve for x.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #22
Solving Applications
Example:
In 2000, a forest had 300 male deer. The growth in the number of
deer is estimated by the function g(t) = 300e0.07t where t is the
number of years after 2000. How many deer will be in the forest in
a) 2010?
b) 2020?
In the year 2000, t = 0.
(Notice that f(0) =300e0.07(0) = 300e0 = 300, the original number of deer.)
In the year 2010, t = 10.
g(10) = 300e0.07(10) = 300e0.7  300(2.01375)  604 deer 2010.
In the year 2020, t = 20.
g(20) = 300e0.07(20) = 300e1.4  300(4.0552)  1217 deer 2010.
Tobey, Slater & Blair, Beginning and Intermediate Algebra, 2e - Slide #23