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

Sections 3.1–3.2Exponential and Logarithmic Functions

V63.0121.021, Calculus I

New York University

October 21, 2010

Announcements

I Midterm is graded and scores are on blackboard. Should get itback in recitation.

I There is WebAssign due Monday/Tuesday next week.

. . . . . .

. . . . . .

Announcements

I Midterm is graded andscores are on blackboard.Should get it back inrecitation.

I There is WebAssign dueMonday/Tuesday nextweek.

V63.0121.021, Calculus I (NYU) Sections 3.1–3.2 Exponential Functions October 21, 2010 2 / 38

. . . . . .

Midterm Statistics

I Average: 78.77%I Median: 80%I Standard Deviation: 12.39%I “good” is anything above average and “great” is anything more

than one standard deviation above average.I More than one SD below the mean is cause for concern.


. . . . . .

Objectives for Sections 3.1 and 3.2

I Know the definition of anexponential function

I Know the properties ofexponential functions

I Understand and apply thelaws of logarithms,including the change ofbase formula.


. . . . . .

Outline

Definition of exponential functions

Properties of exponential Functions

The number e and the natural exponential functionCompound InterestThe number eA limit

Logarithmic Functions


. . . . . .

Derivation of exponential functions

DefinitionIf a is a real number and n is a positive whole number, then

an = a · a · · · · · a︸︷︷︸n factors

Examples

I 23 = 2 · 2 · 2 = 8I 34 = 3 · 3 · 3 · 3 = 81I (−1)5 = (−1)(−1)(−1)(−1)(−1) = −1


. . . . . .

Derivation of exponential functions

DefinitionIf a is a real number and n is a positive whole number, then

an = a · a · · · · · a︸︷︷︸n factors

Examples

I 23 = 2 · 2 · 2 = 8I 34 = 3 · 3 · 3 · 3 = 81I (−1)5 = (−1)(−1)(−1)(−1)(−1) = −1


. . . . . .

Anatomy of a power

DefinitionA power is an expression of the form ab.

I The number a is called the base.I The number b is called the exponent.


. . . . . .

FactIf a is a real number, then

I ax+y = axay (sums to products)

I ax−y =ax

ay

(differences to quotients)

I (ax)y = axy

(repeated exponentiation to multiplied powers)

I (ab)x = axbx

(power of product is product of powers)

whenever all exponents are positive whole numbers.

Proof.Check for yourself:

ax+y = a · a · · · · · a︸︷︷︸x+ y factors

= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .



I ax−y =ax

ay (differences to quotients)

I (ax)y = axy

(repeated exponentiation to multiplied powers)

I (ab)x = axbx





= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .



I ax−y =ax


I (ax)y = axy (repeated exponentiation to multiplied powers)I (ab)x = axbx





= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .



I ax−y =ax


I (ax)y = axy (repeated exponentiation to multiplied powers)I (ab)x = axbx (power of product is product of powers)




= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .



I ax−y =ax






= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .



I ax−y =ax






= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .

Let's be conventional

I The desire that these properties remain true gives us conventionsfor ax when x is not a positive whole number.

I For example, what should a0 be? We cannot write down zero a’sand multiply them together. But we would want this to be true:

an = an+0 != an · a0 =⇒ a0 !

=an

an = 1

(The equality with the exclamation point is what we want.)

DefinitionIf a ̸= 0, we define a0 = 1.

I Notice 00 remains undefined (as a limit form, it’s indeterminate).


. . . . . .




an = an+0 != an · a0

=⇒ a0 !=

an

an = 1





. . . . . .




an = an+0 != an · a0 =⇒ a0 !

=an

an = 1





. . . . . .




an = an+0 != an · a0 =⇒ a0 !

=an

an = 1





. . . . . .




an = an+0 != an · a0 =⇒ a0 !

=an

an = 1





. . . . . .

Conventions for negative exponents

If n ≥ 0, we want

an+(−n) != an · a−n

=⇒ a−n !=

a0

an =1an

Definition

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

Fact

I The convention that a−n =1an “works” for negative n as well.

I If m and n are any integers, then am−n =am

an .


. . . . . .


If n ≥ 0, we want

an+(−n) != an · a−n =⇒ a−n !

=a0

an =1an

Definition


Fact



an .


. . . . . .


If n ≥ 0, we want

an+(−n) != an · a−n =⇒ a−n !

=a0

an =1an

Definition


Fact



an .


. . . . . .


If n ≥ 0, we want

an+(−n) != an · a−n =⇒ a−n !

=a0

an =1an

Definition


Fact



an .


. . . . . .

Conventions for fractional exponents

If q is a positive integer, we want

(a1/q)q != a1 = a

=⇒ a1/q != q

√a

DefinitionIf 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


. . . . . .



(a1/q)q != a1 = a =⇒ a1/q !

= q√a



is even.


ap/q = (ap)1/q = (a1/q)p


. . . . . .



(a1/q)q != a1 = a =⇒ a1/q !

= q√a



is even.


ap/q = (ap)1/q = (a1/q)p


. . . . . .



(a1/q)q != a1 = a =⇒ a1/q !

= q√a



is even.


ap/q = (ap)1/q = (a1/q)p


. . . . . .

Conventions for irrational exponents

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

DefinitionLet a > 0. Then

ax = limr→x

r rational

ar

In other words, to approximate ax for irrational x, take r close to x butrational and compute ar.


. . . . . .




ax = limr→x

r rational

ar



. . . . . .




ax = limr→x

r rational

ar



. . . . . .

Approximating a power with an irrational exponent

r 2r

3 23 = 83.1 231/10 =

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

.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


. . . . . .


. .x

.y

.y = 1x



. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .x

.y

.y = 1x



. . . . . .

Outline






. . . . . .

Properties of exponential Functions.

.

TheoremIf 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 numbersx and y, and positive numbers a and b we have

I ax+y = axay

I ax−y =ax

ay

(negative exponents mean reciprocals)

I (ax)y = axy

(fractional exponents mean roots)

I (ab)x = axbx

Proof.

I This is true for positive integer exponents by natural definitionI Our conventional definitions make these true for rational exponentsI Our limit definition make these for irrational exponents, too


. . . . . .


.


I ax+y = axay

I ax−y =ax

ay (negative exponents mean reciprocals)I (ax)y = axy

(fractional exponents mean roots)

I (ab)x = axbx

Proof.



. . . . . .


.


I ax+y = axay

I ax−y =ax

ay (negative exponents mean reciprocals)I (ax)y = axy (fractional exponents mean roots)I (ab)x = axbx

Proof.



. . . . . .

Simplifying exponential expressions

Example

Simplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:√8

21/2

Answer2


. . . . . .


Example

Simplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:√8

21/2

Answer2


. . . . . .


Example

Simplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:√8

21/2

Answer2


. . . . . .


Example

Simplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:√8

21/2

Answer2


. . . . . .


Example

Simplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:√8

21/2

Answer2V63.0121.021, Calculus I (NYU) Sections 3.1–3.2 Exponential Functions October 21, 2010 17 / 38

. . . . . .

Limits of exponential functions

Fact (Limits of exponentialfunctions)

I If a > 1, then limx→∞

ax = ∞and lim

x→−∞ax = 0

I If 0 < a < 1, thenlimx→∞

ax = 0 andlim

x→−∞ax = ∞ . .x

.y

.y = 1x



. . . . . .

Outline






. . . . . .

Compounded Interest

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have

I After one year?I After two years?I after t years?

Answer

I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.


. . . . . .

Compounded Interest



Answer

I $100+ 10% = $110

I $110+ 10% = $110+ $11 = $121I $100(1.1)t.


. . . . . .

Compounded Interest



Answer

I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121

I $100(1.1)t.


. . . . . .

Compounded Interest



Answer

I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.


. . . . . .

Compounded Interest: quarterly

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have


Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38,

not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!

I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84

I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .

Compounded Interest: monthly

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded twelve times a year. How much do you have after tyears?

Answer$100(1+ 10%/12)12t


. . . . . .

Compounded Interest: monthly

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded twelve times a year. How much do you have after tyears?

Answer$100(1+ 10%/12)12t


. . . . . .

Compounded Interest: general

QuestionSuppose you save P at interest rate r, with interest compounded ntimes a year. How much do you have after t years?

Answer

B(t) = P(1+

rn

)nt


. . . . . .

Compounded Interest: general

QuestionSuppose you save P at interest rate r, with interest compounded ntimes a year. How much do you have after t years?

Answer

B(t) = P(1+

rn

)nt


. . . . . .

Compounded Interest: continuous

QuestionSuppose you save P at interest rate r, with interest compounded everyinstant. How much do you have after t years?

Answer

B(t) = limn→∞

P(1+

rn

)nt= lim

n→∞P(1+

1n

)rnt

= P[

limn→∞

(1+

1n

)n

︸︷︷︸independent of P, r, or t

]rt


. . . . . .

Compounded Interest: continuous

QuestionSuppose you save P at interest rate r, with interest compounded everyinstant. How much do you have after t years?

Answer

B(t) = limn→∞

P(1+

rn

)nt= lim

n→∞P(1+

1n

)rnt

= P[

limn→∞

(1+

1n

)n

︸︷︷︸independent of P, r, or t

]rt


. . . . . .

The magic number

Definition

e = limn→∞

(1+

1n

)n

So now continuously-compounded interest can be expressed as

B(t) = Pert.


. . . . . .

The magic number

Definition

e = limn→∞

(1+

1n

)n

So now continuously-compounded interest can be expressed as

B(t) = Pert.


. . . . . .

Existence of eSee Appendix B

I We can experimentallyverify that this numberexists and is

e ≈ 2.718281828459045 . . .

I e is irrationalI e is transcendental

n(1+

1n

)n

1 22 2.25

3 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.37037

10 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374

100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.70481

1000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692

106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .

I e is irrational

I e is transcendental

n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .

Meet the Mathematician: Leonhard Euler

I Born in Switzerland, livedin Prussia (Germany) andRussia

I Eyesight trouble all his life,blind from 1766 onward

I Hundreds of contributionsto calculus, number theory,graph theory, fluidmechanics, optics, andastronomy

Leonhard Paul EulerSwiss, 1707–1783


. . . . . .

A limit.

.

Question

What is limh→0

eh − 1h

?

Answer

I e = limn→∞

(1+ 1/n)n = limh→0

(1+ h)1/h. So for a small h, e ≈ (1+ h)1/h. So

eh − 1h

≈[(1+ h)1/h

]h − 1h

= 1

I It follows that limh→0

eh − 1h

= 1.

I This can be used to characterize e: limh→0

2h − 1h

= 0.693 · · · < 1 and

limh→0

3h − 1h

= 1.099 · · · > 1


. . . . . .

A limit.

.

Question

What is limh→0

eh − 1h

?

Answer

I e = limn→∞

(1+ 1/n)n = limh→0


eh − 1h

≈[(1+ h)1/h

]h − 1h

= 1


eh − 1h

= 1.


2h − 1h

= 0.693 · · · < 1 and

limh→0

3h − 1h

= 1.099 · · · > 1


. . . . . .

A limit.

.

Question

What is limh→0

eh − 1h

?

Answer

I e = limn→∞

(1+ 1/n)n = limh→0


eh − 1h

≈[(1+ h)1/h

]h − 1h

= 1


eh − 1h

= 1.


2h − 1h

= 0.693 · · · < 1 and

limh→0

3h − 1h

= 1.099 · · · > 1


. . . . . .

Outline






. . . . . .

Logarithms

Definition

I The base a logarithm loga x is the inverse of the function ax

y = loga x ⇐⇒ x = ay

I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.

Facts

(i) loga(x1 · x2) = loga x1 + loga x2

(ii) loga

(x1x2

)= loga x1 − loga x2

(iii) loga(xr) = r loga x


. . . . . .

Logarithms

Definition




Facts


(ii) loga

(x1x2




. . . . . .

Logarithms

Definition




Facts


(ii) loga

(x1x2




. . . . . .

Logarithms

Definition




Facts


(ii) loga

(x1x2




. . . . . .

Logarithms convert products to sums

I Suppose y1 = loga x1 and y2 = loga x2I Then x1 = ay1 and x2 = ay2

I So x1x2 = ay1ay2 = ay1+y2

I Thereforeloga(x1 · x2) = loga x1 + loga x2


. . . . . .

Example

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

Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example

Write as a single logarithm: ln34+ 4 ln 2

Answerln 12


. . . . . .

Example


Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example


Answerln 12


. . . . . .

Example


Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example


Answerln 12


. . . . . .

Example


Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example


Answerln 12


. . . . . .

Graphs of logarithmic functions

. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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

FactIf a > 0 and a ̸= 1, and the same for b, then

loga x =logb xlogb a

Proof.

I If y = loga x, then x = ay

I So logb x = logb(ay) = y logb a

I Thereforey = loga x =

logb xlogb a


. . . . . .

Change of base formula for exponentials

FactIf a > 0 and a ̸= 1, and the same for b, then


Proof.

I If y = loga x, then x = ay

I So logb x = logb(ay) = y logb a

I Thereforey = loga x =

logb xlogb a


. . . . . .

Example of changing base

Example

Find log2 8 by using log10 only.

Solution

log2 8 =log10 8log10 2

≈ 0.903090.30103

= 3

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


. . . . . .


Example


Solution


≈ 0.903090.30103

= 3



. . . . . .


Example


Solution


≈ 0.903090.30103

= 3

Surprised?

No, log2 8 = log2 23 = 3 directly.


. . . . . .


Example


Solution


≈ 0.903090.30103

= 3



. . . . . .

Upshot of changing base

The point of the change of base formula


=1

logb a· logb x = constant · logb x

is that all the logarithmic functions are multiples of each other. So justpick one and call it your favorite.

I Engineers like the common logarithm log = log10I Computer scientists like the binary logarithm lg = log2I Mathematicians like natural logarithm ln = loge

Naturally, we will follow the mathematicians. Just don’t pronounce it“lawn.”


. . . . . .

..“lawn”

.

.Image credit: SelvaV63.0121.021, Calculus I (NYU) Sections 3.1–3.2 Exponential Functions October 21, 2010 37 / 38

http://www.flickr.com/photos/selva/27991

. . . . . .

Summary

I Exponentials turn sums into productsI Logarithms turn products into sumsI Slide rule scabbards are wicked cool


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

Documents

quotientsa ax y

ya ax y

ax ayx

nyu sections

real number

y differences

ax bx power of product

limitlogarithmic functions