CRASH COURSE IN PRECALCULUS
Shiah-Sen Wang
The graphs are prepared by Chien-Lun Lai
Based on : Precalculus: Mathematics for Calculus
by J. Stuwart, L. Redin & S. Watson,
6th edition, 2012, Brooks/Cole
Chapter 4.
LECTURE 5. EXPONENTIAL AND LOGARITHM
FUNCTIONS
The purpose of this lecture is to introduce exponential and
logarithm functions and some of their algebraic properties.
Exponential Functions
For a positive base a ≠ 1, the one-to-one function f (x) = ax is
defined for any exponent x ∈ Q.
Because of great importance in mathematics and in modeling
many real world phenomena, it is desirable to extend the
domain of this one-to-one function to the whole real line R and
1
range (0, ∞), so that f is still one-to-one with a0 = 1, a−x = x
a
and ax+y = ax ⋅ ay for all x, y ∈ R. If such an extension exists,
and it does, its graph in the plane will look like:
Exponential Functions
Note that ax and x a are two different functions. Exercise: How?
Graphs of Exponential Functions for Various Bases
Logarithmic Functions
Let a ≠ 1 be a positive number. The logarithmic function with
base a, denoted by loga ∶ (0, ∞) Ð→ R is defined by
loga x = y
⇔ ay = x
ie., it is the inverse function of the exponential function with
base a.
Using the symmetricity property for the graph of a function and
the graph of its inverse and what we know about the graph of
ax , we have
Logarithmic Functions
Graphs of Logarithmic Functions for Various Bases
In case the base is 10, we simply write log x for log10 x and call
it the common logarithm.
Some Facts about Logarithms
1. loga 1 = 0.
2. loga a = 1.
3. loga ax = x for all x ∈ R.
4. aloga x = x for all x > 0.
Examples
1
1. log49 7 = log49 49 2 =
2. 10log 87 = 87.
1
.
2
Laws of Logarithms
Let a ≠ 1 be a positive number and A, B, C ∈ R with A, B > 0
1. loga (AB) = loga A + loga B⇒ loga (AB) = loga ∣A∣ + loga ∣B∣
whenever A ⋅ B > 0.
A
2. loga ( ) = loga A − loga B⇒ loga ( BA ) = loga ∣A∣ − loga ∣B∣
B
whenever A ⋅ B > 0.
3. loga AC = C loga A.
Examples
1. Use the laws of logarithms to expand
√
1
ab
log ( √ ) = log ab − log c = log ∣a∣ + log ∣b∣ − log c 2
c
log c
= log ∣a∣ + log ∣b∣ −
.
2
2. Use the law of logarithm to write
1
log x − 2 log(x 2 + 1) + log(3 − x 4 ) into a single logarithm.
2
Laws of Logarithms
1
log(3 − x 4 )
2
1
= log x − log(x 2 + 1)2 + log(3 − x 4 ) 2
1
= [log x − log(x 2 + 1)2 ] + log(3 − x 4 ) 2
log x − 2 log(x 2 + 1) +
1
x
x(3 − x 4 ) 2
4 12
= log ( 2
)
+
log(3
−
x
)
=
log
.
(x + 1)2
(x 2 + 1)2
All the above computations make sense ⇔
⎧
x >0
⎪
⎧
⎧
⎪
⎪
⎪
⎪
⎪ 2
⎪x > 0
⎪x > 0
⇔
⎨
⇔
⎨x + 1 > 0 ⇔ ⎨
4
4
⎪
⎪
⎪
3
−
x
>
0
3
>
x
≥
0
⎪
⎪
⎪
⎪
4
⎩
⎩
⎪
⎩3 − x > 0
⎧
√
⎪
⎪x > 0
4
⎨√
⇔
3>x >0
4
⎪
⎪
⎩ 3>x >0
Laws of Logarithms
Here is a list of some reminders about logarithms:
1. loga (A ± B)≠ loga A ± loga B.
A log A
2. loga ( ) ≠ a , unless A = 1.
B loga B
3. loga AC ≠ (loga A)C .
CHANGE OF BASE FORMULA
logb x =
loga x
loga b
for a, b, x > 0 and a, b ≠ 1
⇕
bx = ax loga b
for a, b, x > 0 and a, b ≠ 1
Proof of Change of Base Formula
loga x
for a, b, x > 0 and a, b ≠ 1:
loga b
Starting with the equation y = logb x, we check that
The proof of logb x =
y = logb x ⇐⇒ by = blogb x = x ⇐⇒ loga by = loga x
⇐⇒ y loga b = loga x.
The proof of bx = ax loga b
for a, b, x > 0 and a, b ≠ 1:
Solve the equation bx = ay for y in terms of x.
bx = ay ⇐⇒ loga bx = loga ay ⇐⇒ x loga b = y loga a ⇐⇒ y = x loga b
Examples
Here we give some examples about solving equations involving
exponential and logarithmic functions:
1. 2x+3 = 7.
2x+3 = 7 ⇔ x + 3 = log2 7 ⇔ x = log2 7 − 3.
2. 32x + 3x − 6 = 0.
0 = 32x + 3x − 6 = (3x )2 + 3x − 6 = (3x + 3)(3x − 2) ⇔ 3x + 3 = 0
or 3x = 2 ⇔ 3x = −3 (which is impossible because 3x > 0)
or 3x = 2 ⇔ x = log3 2. Hence the solution is x = log3 2.
3. log(25 − x) = 3.
log(25 − x) = 3 ⇔ 25 − x = 103 = 1000 ⇔ x = −975.
4. log2 (x + 2) + log2 (x − 1) = 1.
log2 (x + 2) + log2 (x − 1) = 1
⇔ log2 [(x + 2)(x − 1)] = 1(by remembering that x > 1)
√
2
17
1 2
2
1
2
) =
⇔ x +x −2 = 2 = 2 ⇔ x +x −4 = 0 ⇔ (x + ) −(
2
√
√
√2
1 + 17
1 − 17
1 ± 17
0 ⇔ (x +
) (x +
)=0⇔x =
.
2
2
2