Lesson 13: Exponential and Logarithmic Functions (Section 041 slides)

Sections 3.1–3.2
Exponential and Logarithmic Functions

V63.0121.041, Calculus I

New York University

October 20, 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.

. . . . . .

V63.0121.041, Calculus I (NYU) Sections 3.1–3.2 Exponential Functions October 20, 2010 2 / 37

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
ax−y = y
a
(ax )y = axy
(ab)x = ax bx
whenever all exponents are positive whole numbers.

. . . . . .


Fact
ax
ax−y = y (differences to quotients)
a
(ax )y = axy
(ab)x = ax bx

. . . . . .


Fact
ax
a
(ax )y = axy (repeated exponentiation to multiplied powers)
(ab)x = ax bx

. . . . . .


Fact
ax
a
(ab)x = ax bx (power of product is product of powers)

. . . . . .


Fact
ax
a
(ab)x = ax bx (power of product is product of powers)

Proof.
Check for yourself:

ax+y = a · a · · · · · a = a · a · · · · · a · a · a · · · · · a = ax ay
x + y factors x factors y factors

. . . . . .


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 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

. . . . . .




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

. . . . . .




! ! an
an = an+0 = an · a0 =⇒ a0 = =1
an

Definition
If a ̸= 0, we define a0 = 1.

. . . . . .




! ! an
an = an+0 = an · a0 =⇒ a0 = =1
an

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

a0 1
an+(−n) = an · a−n =⇒ a−n =
! !
n
= n
a a

. . . . . .


If n ≥ 0, we want

a0 1
an+(−n) = an · a−n =⇒ a−n =
! !
n
= n
a a

Definition
1
If n is a positive integer, we define a−n = .
an

. . . . . .


If n ≥ 0, we want

a0 1
an+(−n) = an · a−n =⇒ a−n =
! !
n
= n
a a

Definition
1
If n is a positive integer, we define a−n = .
an

Fact
1
The convention that a−n = “works” for negative n as well.
an
am
If m and n are any integers, then am−n = n .
a

. . . . . .


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.
√q
( √ )p
Notice that ap = q a . 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?

. . . . . .




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

. . . . . .




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

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
10
3.1 231/10 = √ 31 ≈ 8.57419
2
100
3.14 2314/100 = √2314 ≈ 8.81524
1000
3.141 23141/1000 = 23141 ≈ 8.82135
The limit (numerically approximated is)

2π ≈ 8.82498

. . . . . .


Graphs of various exponential functions
y
.

. x
.
. . . . . .


y
.

. = 1x
y

. x
.
. . . . . .


y
.
. = 2x
y

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. x
.
. . . . . .


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

. = 1x
y

. 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
ax−y = y
a
(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
ax−y = y (negative exponents mean reciprocals)
a
(ax )y = axy
(ab)x = ax bx

Proof.


.
. . . . . .


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

Proof.


.
. . . . . .


Simplifying exponential expressions
Example
Simplify: 82/3

. . . . . .


Example
Simplify: 82/3

Solution
√
3 √
82/3 = 82 =
3
64 = 4

. . . . . .


Example
Simplify: 82/3

Solution
√3 √
82/3 = 82 = 64 = 4
3

( √ )2
8 = 22 = 4.
3
Or,

. . . . . .


Example
Simplify: 82/3

Solution
√3 √
82/3 = 82 = 64 = 4
3

( √ )2
8 = 22 = 4.
3
Or,

Example
√
8
Simplify:
21/2

. . . . . .


Example
Simplify: 82/3

Solution
√3 √
82/3 = 82 = 64 = 4
3

( √ )2
8 = 22 = 4.
3
Or,

Example
√
8
Simplify:
21/2

Answer
2
. . . . . .


Limits of exponential functions

Fact (Limits of exponential y
.
functions) . = (= 2()1/32/3)x
y . 1/ =x( )x
y .
y y y = x . 3x y
. = (. /10)10x= 2x. =
1 . =
y y

If a > 1, then lim ax = ∞
x→∞
and lim ax = 0
x→−∞
If 0 < a < 1, then
lim ax = 0 and y
. =
x→∞
lim a = ∞ x . x
.
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

( r )nt
B(t) = P 1 +
n

. . . . . .


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

( ( )
r )nt 1 rnt
B(t) = lim P 1 + = lim P 1 +
n→∞ n n→∞ n
[ ( )n ]rt
1
=P lim 1 +
n→∞ n
independent of P, r, or t

. . . . . .


The magic number

Definition
( )
1 n
e = lim 1 +
n→∞ n

. . . . . .


The magic number

Definition
( )
1 n
e = lim 1 +
n→∞ n

So now continuously-compounded interest can be expressed as

B(t) = Pert .

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
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

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

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
3 2.37037
e ≈ 2.718281828459045 . . .
10 2.59374
e is irrational 100 2.70481
1000 2.71692
106 2.71828

. . . . . .


Existence of e
See Appendix B

( )
1 n
n 1+
3 2.37037
e ≈ 2.718281828459045 . . .
10 2.59374
e is irrational 100 2.70481
1000 2.71692
e is transcendental
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
eh − 1
What is lim ?
h→0 h

. . . . . . .


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

Answer

e = lim → ∞ (1 + 1/n)n = lim (1 + h)1/h . So for a small h, e ≈ (1 + h)1/h .
n h→0
So [ ]h
eh − 1 (1 + h)1/h − 1
≈ =1
h h

. . . . . . .


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

Answer

e = lim → ∞ (1 + 1/n)n = lim (1 + h)1/h . So for a small h, e ≈ (1 + h)1/h .
n h→0
So [ ]h
eh − 1 (1 + h)1/h − 1
≈ =1
h h

eh − 1
It follows that lim = 1.
h→0 h
2h − 1
This can be used to characterize e: lim = 0.693 · · · < 1 and
h→0 h
3h − 1
lim = 1.099 · · · > 1
h→0 h
. . . . . . .


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



y = ln x ⇐⇒ x = ey .

Facts

(i) loga (x1 · x2 ) = loga x1 + loga x2

. . . . . .


Logarithms

Definition



y = ln x ⇐⇒ x = ey .

Facts

( )
x1
(ii) loga = loga x1 − loga x2
x2

. . . . . .


Logarithms

Definition



y = ln x ⇐⇒ x = ey .

Facts

( )
x1
(ii) loga = loga x1 − loga x2
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 x1 x2 = 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
42
2 ln 4 − ln 3 = ln 42 − ln 3 = ln
3
ln 42
not !
ln 3

. . . . . .


Example

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

Example
3
Write as a single logarithm: ln + 4 ln 2
4

. . . . . .


Example

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

Example
3
Write as a single logarithm: ln + 4 ln 2
4

Answer
ln 12
. . . . . .


Graphs of logarithmic functions
y
.
. = 2x
y

y
. = log2 x

. . 0, 1)
(

..1, 0) .
( x
.

. . . . . .


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

y
. = log2 x

y
. = log3 x
. . 0, 1)
(

..1, 0) .
( x
.

. . . . . .


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

y
. = log2 x

y
. = log3 x
. . 0, 1)
(
y
. = log10 x
..1, 0) .
( x
.

. . . . . .


y
.
. = .10=3x= 2x
y xy
y y. = .ex

y
. = log2 x

y
. = ln x
y
. = log3 x
. . 0, 1)
(
y
. = log10 x
..1, 0) .
( x
.

. . . . . .


Change of base formula for exponentials

Fact
If a > 0 and a ̸= 1, then
ln x
loga x =
ln a

. . . . . .


Change of base formula for exponentials

Fact
If a > 0 and a ̸= 1, then
ln x
loga x =
ln a

Proof.

If y = loga x, then x = ay
So ln x = ln(ay ) = y ln a
Therefore
ln x
y = loga x =
ln a

. . . . . .


Example of changing base

Example
Find log2 8 by using log10 only.

Surprised?

. . . . . .



Example

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

Surprised?

. . . . . .



Example

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

Surprised? No, log2 8 = log2 23 = 3 directly.

. . . . . .


Upshot of changing base

The point of the change of base formula

logb x 1
loga x = = · logb x = constant · logb x
logb a logb a

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

. . . . . .


“ .
. 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 041 slides)

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