Lesson 13: Exponential and Logarithmic Functions (Section 021 slides)

Sections 3.1–3.2
Exponential and Logarithmic Functions
V63.0121.021, Calculus I
New York University
October 21, 2010
Announcements
Midterm is graded and scores are on blackboard. Should get it
back in recitation.
There is WebAssign due Monday/Tuesday next week.
. . . . . .

. . . . . .
Announcements
Midterm is graded and
scores are on blackboard.
Should get it back in
recitation.
There is WebAssign due
Monday/Tuesday next
week.
V63.0121.021, Calculus I (NYU) Sections 3.1–3.2 Exponential Functions October 21, 2010 2 / 38

. . . . . .
Midterm Statistics
Average: 78.77%
Median: 80%
Standard Deviation: 12.39%
“good” is anything above average and “great” is anything more
than one standard deviation above average.
More than one SD below the mean is cause for concern.

. . . . . .
Objectives for Sections 3.1 and 3.2
Know the definition of an
exponential function
Know the properties of
exponential functions
Understand and apply the
laws of logarithms,
including the change of
base formula.

. . . . . .
Outline
Definition of exponential functions
Properties of exponential Functions
The number e and the natural exponential function
Compound Interest
The number e
A limit
Logarithmic Functions

. . . . . .
Derivation of exponential functions
Definition
If a is a real number and n is a positive whole number, then
an
= a · a · · · · · a
n factors

. . . . . .
Derivation of exponential functions
Definition
If a is a real number and n is a positive whole number, then
an
= a · a · · · · · a
n factors
Examples
23
= 2 · 2 · 2 = 8
34
= 3 · 3 · 3 · 3 = 81
(−1)5
= (−1)(−1)(−1)(−1)(−1) = −1

. . . . . .
Anatomy of a power
Definition
A power is an expression of the form ab
.
The number a is called the base.
The number b is called the exponent.

. . . . . .
Fact
If a is a real number, then
ax+y
= ax
ay
(sums to products)
ax−y
=
ax
ay
(ax
)y
= axy
(ab)x
= ax
bx
whenever all exponents are positive whole numbers.

. . . . . .
Fact
ax+y
= ax
ay
(sums to products)
ax−y
=
ax
ay
(differences to quotients)
(ax
)y
= axy
(ab)x
= ax
bx

. . . . . .
Fact
ax+y
= ax
ay
(sums to products)
ax−y
=
ax
ay
(ax
)y
= axy
(repeated exponentiation to multiplied powers)
(ab)x
= ax
bx

. . . . . .
Fact
ax+y
= ax
ay
(sums to products)
ax−y
=
ax
ay
(ax
)y
= axy
(ab)x
= ax
bx
(power of product is product of powers)

. . . . . .
Fact
ax+y
= ax
ay
(sums to products)
ax−y
=
ax
ay
(ax
)y
= axy
(ab)x
= ax
bx
(power of product is product of powers)
Proof.
Check for yourself:
ax+y
= a · a · · · · · a
x + y factors
= a · a · · · · · a
x factors
· a · a · · · · · a
y factors
= ax
ay

. . . . . .
Let's be conventional
The desire that these properties remain true gives us conventions
for ax
when x is not a positive whole number.

. . . . . .
for ax
For example, what should a0
be? We cannot write down zero a’s
and multiply them together. But we would want this to be true:
an
= an+0 !
= an
· a0

. . . . . .
for ax
an
= an+0 !
= an
· a0
=⇒ a0 !
=
an
an
= 1
(The equality with the exclamation point is what we want.)

. . . . . .
for ax
an
= an+0 !
= an
· a0
=⇒ a0 !
=
an
an
= 1
Definition
If a ̸= 0, we define a0
= 1.

. . . . . .
for ax
an
= an+0 !
= an
· a0
=⇒ a0 !
=
an
an
= 1
Definition
If a ̸= 0, we define a0
= 1.
Notice 00
remains undefined (as a limit form, it’s indeterminate).

. . . . . .
Conventions for negative exponents
If n ≥ 0, we want
an+(−n) !
= an
· a−n

. . . . . .
If n ≥ 0, we want
an+(−n) !
= an
· a−n
=⇒ a−n !
=
a0
an
=
1
an

. . . . . .
If n ≥ 0, we want
an+(−n) !
= an
· a−n
=⇒ a−n !
=
a0
an
=
1
an
Definition
If n is a positive integer, we define a−n
=
1
an
.

. . . . . .
If n ≥ 0, we want
an+(−n) !
= an
· a−n
=⇒ a−n !
=
a0
an
=
1
an
Definition
If n is a positive integer, we define a−n
=
1
an
.
Fact
The convention that a−n
=
1
an
“works” for negative n as well.
If m and n are any integers, then am−n
=
am
an
.

. . . . . .
Conventions for fractional exponents
If q is a positive integer, we want
(a1/q
)q !
= a1
= a

. . . . . .
(a1/q
)q !
= a1
= a =⇒ a1/q !
= q
√
a

. . . . . .
(a1/q
)q !
= a1
= a =⇒ a1/q !
= q
√
a
Definition
If q is a positive integer, we define a1/q
= q
√
a. We must have a ≥ 0 if q
is even.

. . . . . .
(a1/q
)q !
= a1
= a =⇒ a1/q !
= q
√
a
Definition
If q is a positive integer, we define a1/q
= q
√
a. We must have a ≥ 0 if q
is even.
Notice that
q
√
ap =
( q
√
a
)p
. So we can unambiguously say
ap/q
= (ap
)1/q
= (a1/q
)p

. . . . . .
Conventions for irrational exponents
So ax
is well-defined if a is positive and x is rational.
What about irrational powers?

. . . . . .
So ax
Definition
Let a > 0. Then
ax
= lim
r→x
r rational
ar

. . . . . .
So ax
Definition
Let a > 0. Then
ax
= lim
r→x
r rational
ar
In other words, to approximate ax
for irrational x, take r close to x but
rational and compute ar
.

. . . . . .
Approximating a power with an irrational exponent
r 2r
3 23
= 8
3.1 231/10
=
10
√
231
≈ 8.57419
3.14 2314/100
=
100
√
2314
≈ 8.81524
3.141 23141/1000
=
1000
√
23141
≈ 8.82135
The limit (numerically approximated is)
2π
≈ 8.82498

. . . . . .
Graphs of various exponential functions
. .x
.y

. . . . . .
. .x
.y
.y = 1x

. . . . . .
. .x
.y
.y = 1x
.y = 2x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x
.y = 1.5x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x
.y = 1.5x
.y = (1/2)x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x
.y = 1.5x
.y = (1/2)x.y = (1/3)x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x
.y = 1.5x
.y = (1/2)x.y = (1/3)x
.y = (1/10)x

. . . . . .
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x
.y = 1.5x
.y = (1/2)x.y = (1/3)x
.y = (1/10)x
.y = (2/3)x

. . . . . .
Outline
Compound Interest
The number e
A limit

. . . . . .
.
.
Theorem
If a > 0 and a ̸= 1, then f(x) = ax
is a continuous function with domain
(−∞, ∞) and range (0, ∞). In particular, ax
> 0 for all x. For any real numbers
x and y, and positive numbers a and b we have
ax+y
= ax
ay
ax−y
=
ax
ay
(ax
)y
= axy
(ab)x
= ax
bx
Proof.
This is true for positive integer exponents by natural definition
Our conventional definitions make these true for rational exponents
Our limit definition make these for irrational exponents, too

. . . . . .
.
.
Theorem
ax+y
= ax
ay
ax−y
=
ax
ay
(negative exponents mean reciprocals)
(ax
)y
= axy
(ab)x
= ax
bx
Proof.

. . . . . .
.
.
Theorem
ax+y
= ax
ay
ax−y
=
ax
ay
(negative exponents mean reciprocals)
(ax
)y
= axy
(fractional exponents mean roots)
(ab)x
= ax
bx
Proof.

. . . . . .
Simplifying exponential expressions
Example
Simplify: 82/3

. . . . . .
Example
Simplify: 82/3
Solution
82/3
=
3
√
82
=
3
√
64 = 4

. . . . . .
Example
Simplify: 82/3
Solution
82/3
=
3
√
82
=
3
√
64 = 4
Or,
(
3
√
8
)2
= 22
= 4.

. . . . . .
Example
Simplify: 82/3
Solution
82/3
=
3
√
82
=
3
√
64 = 4
Or,
(
3
√
8
)2
= 22
= 4.
Example
Simplify:
√
8
21/2

. . . . . .
Example
Simplify: 82/3
Solution
82/3
=
3
√
82
=
3
√
64 = 4
Or,
(
3
√
8
)2
= 22
= 4.
Example
Simplify:
√
8
21/2
Answer
2

. . . . . .
Limits of exponential functions
Fact (Limits of exponential
functions)
If a > 1, then lim
x→∞
ax
= ∞
and lim
x→−∞
ax
= 0
If 0 < a < 1, then
lim
x→∞
ax
= 0 and
lim
x→−∞
ax
= ∞ . .x
.y
.y =
.y = 2x
.y = 3x
.y = 10x
.y =.y = (1/2)x.y = (1/3)x
.y = (1/10)x
.y = (2/3)x

. . . . . .
Outline
Compound Interest
The number e
A limit

. . . . . .
Compounded Interest
Question
Suppose you save $100 at 10% annual interest, with interest
compounded once a year. How much do you have
After one year?
After two years?
after t years?

. . . . . .
Compounded Interest
Question
After one year?
After two years?
after t years?
Answer
$100 + 10% = $110

. . . . . .
Compounded Interest
Question
After one year?
After two years?
after t years?
Answer
$100 + 10% = $110
$110 + 10% = $110 + $11 = $121

. . . . . .
Compounded Interest
Question
After one year?
After two years?
after t years?
Answer
$100 + 10% = $110
$110 + 10% = $110 + $11 = $121
$100(1.1)t
.

. . . . . .
Compounded Interest: quarterly
Question
compounded four times a year. How much do you have
After one year?
After two years?
after t years?

. . . . . .
Question
After one year?
After two years?
after t years?
Answer
$100(1.025)4
= $110.38,

. . . . . .
Question
After one year?
After two years?
after t years?
Answer
$100(1.025)4
= $110.38, not $100(1.1)4
!

. . . . . .
Question
After one year?
After two years?
after t years?
Answer
$100(1.025)4
= $110.38, not $100(1.1)4
!
$100(1.025)8
= $121.84

. . . . . .
Question
After one year?
After two years?
after t years?
Answer
$100(1.025)4
= $110.38, not $100(1.1)4
!
$100(1.025)8
= $121.84
$100(1.025)4t
.

. . . . . .
Compounded Interest: monthly
Question
compounded twelve times a year. How much do you have after t
years?

. . . . . .
Compounded Interest: monthly
Question
compounded twelve times a year. How much do you have after t
years?
Answer
$100(1 + 10%/12)12t

. . . . . .
Compounded Interest: general
Question
Suppose you save P at interest rate r, with interest compounded n
times a year. How much do you have after t years?

. . . . . .
Compounded Interest: general
Question
Suppose you save P at interest rate r, with interest compounded n
times a year. How much do you have after t years?
Answer
B(t) = P
(
1 +
r
n
)nt

. . . . . .
Compounded Interest: continuous
Question
Suppose you save P at interest rate r, with interest compounded every
instant. How much do you have after t years?

. . . . . .
Compounded Interest: continuous
Question
Suppose you save P at interest rate r, with interest compounded every
instant. How much do you have after t years?
Answer
B(t) = lim
n→∞
P
(
1 +
r
n
)nt
= lim
n→∞
P
(
1 +
1
n
)rnt
= P
[
lim
n→∞
(
1 +
1
n
)n
independent of P, r, or t
]rt

. . . . . .
The magic number
Definition
e = lim
n→∞
(
1 +
1
n
)n

. . . . . .
The magic number
Definition
e = lim
n→∞
(
1 +
1
n
)n
So now continuously-compounded interest can be expressed as
B(t) = Pert
.

. . . . . .
Existence of e
See Appendix B
n
(
1 +
1
n
)n
1 2
2 2.25

. . . . . .
Existence of e
See Appendix B
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037

. . . . . .
Existence of e
See Appendix B
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374

. . . . . .
Existence of e
See Appendix B
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481

. . . . . .
Existence of e
See Appendix B
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692

. . . . . .
Existence of e
See Appendix B
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106
2.71828

. . . . . .
Existence of e
See Appendix B
We can experimentally
verify that this number
exists and is
e ≈ 2.718281828459045 . . .
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106
2.71828

. . . . . .
Existence of e
See Appendix B
exists and is
e ≈ 2.718281828459045 . . .
e is irrational
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106
2.71828

. . . . . .
Existence of e
See Appendix B
exists and is
e ≈ 2.718281828459045 . . .
e is irrational
e is transcendental
n
(
1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106
2.71828

. . . . . .
Meet the Mathematician: Leonhard Euler
Born in Switzerland, lived
in Prussia (Germany) and
Russia
Eyesight trouble all his life,
blind from 1766 onward
Hundreds of contributions
to calculus, number theory,
graph theory, fluid
mechanics, optics, and
astronomy
Leonhard Paul Euler
Swiss, 1707–1783

. . . . . .
A limit
.
.
Question
What is lim
h→0
eh − 1
h
?

. . . . . .
A limit
.
.
Question
What is lim
h→0
eh − 1
h
?
Answer
e = lim
n→∞
(1 + 1/n)n
= lim
h→0
(1 + h)1/h
. So for a small h, e ≈ (1 + h)1/h
. So
eh − 1
h
≈
[
(1 + h)1/h
]h
− 1
h
= 1

. . . . . .
A limit
.
.
Question
What is lim
h→0
eh − 1
h
?
Answer
e = lim
n→∞
(1 + 1/n)n
= lim
h→0
(1 + h)1/h
. So for a small h, e ≈ (1 + h)1/h
. So
eh − 1
h
≈
[
(1 + h)1/h
]h
− 1
h
= 1
It follows that lim
h→0
eh − 1
h
= 1.
This can be used to characterize e: lim
h→0
2h
− 1
h
= 0.693 · · · < 1 and
lim
h→0
3h
− 1
h
= 1.099 · · · > 1

. . . . . .
Outline
Compound Interest
The number e
A limit

. . . . . .
Logarithms
Definition
The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex
. So
y = ln x ⇐⇒ x = ey
.

. . . . . .
Logarithms
Definition
. So
y = ln x ⇐⇒ x = ey
.
Facts
(i) loga(x1 · x2) = loga x1 + loga x2

. . . . . .
Logarithms
Definition
. So
y = ln x ⇐⇒ x = ey
.
Facts
(ii) loga
(
x1
x2
)
= loga x1 − loga x2

. . . . . .
Logarithms
Definition
. So
y = ln x ⇐⇒ x = ey
.
Facts
(ii) loga
(
x1
x2
)
= loga x1 − loga x2
(iii) loga(xr
) = r loga x

. . . . . .
Logarithms convert products to sums
Suppose y1 = loga x1 and y2 = loga x2
Then x1 = ay1 and x2 = ay2
So x1x2 = ay1 ay2 = ay1+y2
Therefore
loga(x1 · x2) = loga x1 + loga x2

. . . . . .
Example
Write as a single logarithm: 2 ln 4 − ln 3.

. . . . . .
Example
Solution
2 ln 4 − ln 3 = ln 42
− ln 3 = ln
42
3
not
ln 42
ln 3
!

. . . . . .
Example
Solution
2 ln 4 − ln 3 = ln 42
− ln 3 = ln
42
3
not
ln 42
ln 3
!
Example
Write as a single logarithm: ln
3
4
+ 4 ln 2

. . . . . .
Example
Solution
2 ln 4 − ln 3 = ln 42
− ln 3 = ln
42
3
not
ln 42
ln 3
!
Example
Write as a single logarithm: ln
3
4
+ 4 ln 2
Answer
ln 12

. . . . . .
Graphs of logarithmic functions
. .x
.y
.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)

. . . . . .
. .x
.y
.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x

. . . . . .
. .x
.y
.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x
.y = 10x
.y = log10 x

. . . . . .
. .x
.y
.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x
.y = 10x
.y = log10 x
.y = ex
.y = ln x

. . . . . .
Change of base formula for exponentials
Fact
If a > 0 and a ̸= 1, and the same for b, then
loga x =
logb x
logb a

. . . . . .
Change of base formula for exponentials
Fact
If a > 0 and a ̸= 1, and the same for b, then
loga x =
logb x
logb a
Proof.
If y = loga x, then x = ay
So logb x = logb(ay
) = y logb a
Therefore
y = loga x =
logb x
logb a

. . . . . .
Example of changing base
Example
Find log2 8 by using log10 only.

. . . . . .
Example
Solution
log2 8 =
log10 8
log10 2
≈
0.90309
0.30103
= 3

. . . . . .
Example
Solution
log2 8 =
log10 8
log10 2
≈
0.90309
0.30103
= 3
Surprised?

. . . . . .
Example
Solution
log2 8 =
log10 8
log10 2
≈
0.90309
0.30103
= 3
Surprised? No, log2 8 = log2 23
= 3 directly.

. . . . . .
Upshot of changing base
The point of the change of base formula
loga x =
logb x
logb a
=
1
logb a
· logb x = constant · logb x
is that all the logarithmic functions are multiples of each other. So just
pick one and call it your favorite.
Engineers like the common logarithm log = log10
Computer scientists like the binary logarithm lg = log2
Mathematicians like natural logarithm ln = loge
Naturally, we will follow the mathematicians. Just don’t pronounce it
“lawn.”

. . . . . .
.
.“lawn”
.
.Image credit: Selva

. . . . . .
Summary
Exponentials turn sums into products
Logarithms turn products into sums
Slide rule scabbards are wicked cool

Lesson 13: Exponential and Logarithmic Functions (Section 021 slides)

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