lesson 13: exponential and logarithmic functions (slides)
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Sec on 3.1–3.2Exponen al and Logarithmic
Func ons
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
March 9, 2011
Announcements
I Midterm is graded.average = 44, median=46,SD =10
I There is WebAssign duea er Spring Break.
I Quiz 3 on 2.6, 2.8, 3.1, 3.2on March 30
Midterm Statistics
I Average: 43.86/60 = 73.1%I Median: 46/60 = 76.67%I Standard Devia on: 10.64%I “good” is anything above average and “great” is anything morethan one standard devia on 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 defini on of anexponen al func on
I Know the proper es ofexponen al func ons
I Understand and applythe laws of logarithms,including the change ofbase formula.
OutlineDefini on of exponen al func ons
Proper es of exponen al Func ons
The number e and the natural exponen al func onCompound InterestThe number eA limit
Logarithmic Func ons
Derivation of exponentialsDefini onIf a is a real number and n is a posi ve 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 exponentialsDefini onIf a is a real number and n is a posi ve 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
Defini onA 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 quo ents)
I (ax)y = axy
(repeated exponen a on to mul plied powers)
I (ab)x = axbx
(power of product is product of powers)
whenever all exponents are posi ve 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
FactIf a is a real number, then
I ax+y = axay (sums to products)
I ax−y =ax
ay (differences to quo ents)
I (ax)y = axy
(repeated exponen a on to mul plied powers)
I (ab)x = axbx
(power of product is product of powers)
whenever all exponents are posi ve 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
FactIf a is a real number, then
I ax+y = axay (sums to products)
I ax−y =ax
ay (differences to quo ents)
I (ax)y = axy (repeated exponen a on to mul plied powers)I (ab)x = axbx
(power of product is product of powers)
whenever all exponents are posi ve 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
FactIf a is a real number, then
I ax+y = axay (sums to products)
I ax−y =ax
ay (differences to quo ents)
I (ax)y = axy (repeated exponen a on to mul plied powers)I (ab)x = axbx (power of product is product of powers)
whenever all exponents are posi ve 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
FactIf a is a real number, then
I ax+y = axay (sums to products)
I ax−y =ax
ay (differences to quo ents)
I (ax)y = axy (repeated exponen a on to mul plied powers)I (ab)x = axbx (power of product is product of powers)
whenever all exponents are posi ve 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
FactIf a is a real number, then
I ax+y = axay (sums to products)
I ax−y =ax
ay (differences to quo ents)
I (ax)y = axy (repeated exponen a on to mul plied powers)I (ab)x = axbx (power of product is product of powers)
whenever all exponents are posi ve 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
Let’s be conventionalI The desire that these proper es remain true gives usconven ons for ax when x is not a posi ve whole number.
I For example, what should a0 be?We would want this to be true:
an = an+0 != an · a0 =⇒ a0 !
=an
an = 1
Defini onIf a ̸= 0, we define a0 = 1.
I No ce 00 remains undefined (as a limit form, it’sindeterminate).
Let’s be conventionalI The desire that these proper es remain true gives usconven ons for ax when x is not a posi ve whole number.
I For example, what should a0 be?We would want this to be true:
an = an+0 != an · a0
=⇒ a0 !=
an
an = 1
Defini onIf a ̸= 0, we define a0 = 1.
I No ce 00 remains undefined (as a limit form, it’sindeterminate).
Let’s be conventionalI The desire that these proper es remain true gives usconven ons for ax when x is not a posi ve whole number.
I For example, what should a0 be?We would want this to be true:
an = an+0 != an · a0 =⇒ a0 !
=an
an = 1
Defini onIf a ̸= 0, we define a0 = 1.
I No ce 00 remains undefined (as a limit form, it’sindeterminate).
Let’s be conventionalI The desire that these proper es remain true gives usconven ons for ax when x is not a posi ve whole number.
I For example, what should a0 be?We would want this to be true:
an = an+0 != an · a0 =⇒ a0 !
=an
an = 1
Defini onIf a ̸= 0, we define a0 = 1.
I No ce 00 remains undefined (as a limit form, it’sindeterminate).
Let’s be conventionalI The desire that these proper es remain true gives usconven ons for ax when x is not a posi ve whole number.
I For example, what should a0 be?We would want this to be true:
an = an+0 != an · a0 =⇒ a0 !
=an
an = 1
Defini onIf a ̸= 0, we define a0 = 1.
I No ce 00 remains undefined (as a limit form, it’sindeterminate).
Conventions for negative exponents
If n ≥ 0, we want
an+(−n) != an · a−n
=⇒ a−n !=
a0
an =1an
Defini on
If n is a posi ve integer, we define a−n =1an .
Conventions for negative exponents
If n ≥ 0, we want
an+(−n) != an · a−n =⇒ a−n !
=a0
an =1an
Defini on
If n is a posi ve integer, we define a−n =1an .
Conventions for negative exponents
If n ≥ 0, we want
an+(−n) != an · a−n =⇒ a−n !
=a0
an =1an
Defini on
If n is a posi ve integer, we define a−n =1an .
Defini on
If n is a posi ve integer, we define a−n =1an .
Fact
I The conven on that a−n =1an “works” for nega ve n as well.
I If m and n are any integers, then am−n =am
an .
Defini on
If n is a posi ve integer, we define a−n =1an .
Fact
I The conven on that a−n =1an “works” for nega ve n as well.
I If m and n are any integers, then am−n =am
an .
Conventions for fractional exponentsIf q is a posi ve integer, we want
(a1/q)q != a1 = a
=⇒ a1/q != q
√a
Defini onIf q is a posi ve integer, we define a1/q = q
√a. We must have a ≥ 0
if q is even.
No ce that q√ap =
(q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
Conventions for fractional exponentsIf q is a posi ve integer, we want
(a1/q)q != a1 = a =⇒ a1/q !
= q√a
Defini onIf q is a posi ve integer, we define a1/q = q
√a. We must have a ≥ 0
if q is even.
No ce that q√ap =
(q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
Conventions for fractional exponentsIf q is a posi ve integer, we want
(a1/q)q != a1 = a =⇒ a1/q !
= q√a
Defini onIf q is a posi ve integer, we define a1/q = q
√a. We must have a ≥ 0
if q is even.
No ce that q√ap =
(q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
Conventions for fractional exponentsIf q is a posi ve integer, we want
(a1/q)q != a1 = a =⇒ a1/q !
= q√a
Defini onIf q is a posi ve integer, we define a1/q = q
√a. We must have a ≥ 0
if q is even.
No ce that q√ap =
(q√a)p. So we can unambiguously say
ap/q = (ap)1/q = (a1/q)p
Conventions for irrationalexponents
I So ax is well-defined if a is posi ve and x is ra onal.I What about irra onal powers?
Defini onLet a > 0. Then
ax = limr→x
r ra onalar
In other words, to approximate ax for irra onal x, take r close to xbut ra onal and compute ar.
Conventions for irrationalexponents
I So ax is well-defined if a is posi ve and x is ra onal.I What about irra onal powers?
Defini onLet a > 0. Then
ax = limr→x
r ra onalar
In other words, to approximate ax for irra onal x, take r close to xbut ra onal and compute ar.
Conventions for irrationalexponents
I So ax is well-defined if a is posi ve and x is ra onal.I What about irra onal powers?
Defini onLet a > 0. Then
ax = limr→x
r ra onalar
In other words, to approximate ax for irra onal x, take r close to xbut ra onal and compute ar.
Approximating a power with anirrational exponent
r 2r
3 23 = 83.1 231/10 = 10
√231 ≈ 8.57419
3.14 2314/100 = 100√
2314 ≈ 8.815243.141 23141/1000 = 1000
√23141 ≈ 8.82135
The limit (numerically approximated is)
2π ≈ 8.82498
Graphs of exponential functions
.. x.
y
Graphs of exponential functions
.. x.
y
.y = 1x
Graphs of exponential functions
.. x.
y
.y = 1x
.
y = 2x
Graphs of exponential functions
.. x.
y
.y = 1x
.
y = 2x
.
y = 3x
Graphs of exponential functions
.. x.
y
.y = 1x
.
y = 2x
.
y = 3x
.
y = 10x
Graphs of exponential functions
.. x.
y
.y = 1x
.
y = 2x
.
y = 3x
.
y = 10x
.
y = 1.5x
Graphs of exponential functions
.. x.
y
.y = 1x
.
y = 2x
.
y = 3x
.
y = 10x
.
y = 1.5x
.
y = (1/2)x
Graphs of exponential functions
.. x.
y
.y = 1x
.
y = 2x
.
y = 3x
.
y = 10x
.
y = 1.5x
.
y = (1/2)x
.
y = (1/3)x
Graphs of 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
Graphs of 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
OutlineDefini on of exponen al func ons
Proper es of exponen al Func ons
The number e and the natural exponen al func onCompound InterestThe number eA limit
Logarithmic Func ons
Properties of exponential Functions
TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have
I ax+y = axay
I ax−y =ax
ay
(nega ve exponents mean reciprocals)
I (ax)y = axy
(frac onal exponents mean roots)
I (ab)x = axbx
Properties of exponential Functions
TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have
I ax+y = axay
I ax−y =ax
ay (nega ve exponents mean reciprocals)
I (ax)y = axy
(frac onal exponents mean roots)
I (ab)x = axbx
Properties of exponential Functions
TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have
I ax+y = axay
I ax−y =ax
ay (nega ve exponents mean reciprocals)
I (ax)y = axy (frac onal exponents mean roots)I (ab)x = axbx
Proof.
I This is true for posi ve integer exponents by natural defini onI Our conven onal defini ons make these true for ra onalexponents
I Our limit defini on make these for irra onal exponents, too
Simplifying exponential expressions
Example
Simplify: 82/3
Solu on
I 82/3 = 3√
82 = 3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Simplifying exponential expressions
Example
Simplify: 82/3
Solu on
I 82/3 = 3√
82 = 3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Simplifying exponential expressions
Example
Simplify: 82/3
Solu on
I 82/3 = 3√
82 = 3√64 = 4
I Or,(
3√8)2
= 22 = 4.
Simplifying exponentialexpressions
Example
Simplify:√8
21/2
Answer2
Simplifying exponentialexpressions
Example
Simplify:√8
21/2
Answer2
Limits of exponential functionsFact (Limits of exponen alfunc ons)
I If a > 1, thenlimx→∞
ax = ∞ andlim
x→−∞ax = 0
I If 0 < a < 1, thenlimx→∞
ax = 0 andlim
x→−∞ax = ∞
.. 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
OutlineDefini on of exponen al func ons
Proper es of exponen al Func ons
The number e and the natural exponen al func onCompound InterestThe number eA limit
Logarithmic Func ons
Compounded InterestQues on
Suppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have A er one year?A er two years? A er t years?
Answer
I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.
Compounded InterestQues on
Suppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have A er one year?A er two years? A er t years?
Answer
I $100+ 10% = $110
I $110+ 10% = $110+ $11 = $121I $100(1.1)t.
Compounded InterestQues on
Suppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have A er one year?A er two years? A er t years?
Answer
I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121
I $100(1.1)t.
Compounded InterestQues on
Suppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have A er one year?A er two years? A er t years?
Answer
I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.
Compounded Interest: quarterlyQues on
Suppose you save $100 at 10% annual interest, with interestcompounded four mes a year. How much do you have A er oneyear? A er two years? A er t years?
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: quarterlyQues on
Suppose you save $100 at 10% annual interest, with interestcompounded four mes a year. How much do you have A er oneyear? A er two years? A er t years?
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: quarterlyQues on
Suppose you save $100 at 10% annual interest, with interestcompounded four mes a year. How much do you have A er oneyear? A er two years? A er t years?
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: quarterlyQues on
Suppose you save $100 at 10% annual interest, with interestcompounded four mes a year. How much do you have A er oneyear? A er two years? A er t years?
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.
Compounded Interest: quarterlyQues on
Suppose you save $100 at 10% annual interest, with interestcompounded four mes a year. How much do you have A er oneyear? A er two years? A er t years?
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
Ques on
Suppose you save $100 at 10% annual interest, with interestcompounded twelve mes a year. How much do you have a er tyears?
Answer$100(1+ 10%/12)12t
Compounded Interest: monthly
Ques on
Suppose you save $100 at 10% annual interest, with interestcompounded twelve mes a year. How much do you have a er tyears?
Answer$100(1+ 10%/12)12t
Compounded Interest: general
Ques on
Suppose you save P at interest rate r, with interest compounded nmes a year. How much do you have a er t years?
Answer
B(t) = P(1+
rn
)nt
Compounded Interest: general
Ques on
Suppose you save P at interest rate r, with interest compounded nmes a year. How much do you have a er t years?
Answer
B(t) = P(1+
rn
)nt
Compounded Interest: continuousQues on
Suppose you save P at interest rate r, with interest compoundedevery instant. How much do you have a er 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: continuousQues on
Suppose you save P at interest rate r, with interest compoundedevery instant. How much do you have a er 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 numberDefini on
e = limn→∞
(1+
1n
)n
So now con nuously-compounded interest can be expressed as
B(t) = Pert.
The magic numberDefini on
e = limn→∞
(1+
1n
)n
So now con nuously-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 irra onalI e is transcendental
n(1+
1n
)n
1 22 2.25
3 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.37037
10 2.59374100 2.704811000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374
100 2.704811000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.70481
1000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692
106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onal
I e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Meet the Mathematician: Leonhard EulerI Born in Switzerland, livedin Prussia (Germany) andRussia
I Eyesight trouble all hislife, blind from 1766onward
I Hundreds ofcontribu ons to calculus,number theory, graphtheory, fluid mechanics,op cs, and astronomy
Leonhard Paul EulerSwiss, 1707–1783
A limitQues on
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
A limitQues on
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
A limit
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
OutlineDefini on of exponen al func ons
Proper es of exponen al Func ons
The number e and the natural exponen al func onCompound InterestThe number eA limit
Logarithmic Func ons
Logarithms
Defini on
I The base a logarithm loga x is the inverse of the func on ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.
Facts about Logarithms
Facts
(i) loga(x1 · x2) = loga x1 + loga x2
(ii) loga
(x1x2
)= loga x1 − loga x2
(iii) loga(xr) = r loga x
Facts about Logarithms
Facts
(i) loga(x1 · x2) = loga x1 + loga x2
(ii) loga
(x1x2
)= loga x1 − loga x2
(iii) loga(xr) = r loga x
Facts about Logarithms
Facts
(i) loga(x1 · x2) = loga x1 + loga x2
(ii) loga
(x1x2
)= loga x1 − loga x2
(iii) loga(xr) = r loga x
Logarithms convert products to sumsI 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
ExamplesExample
Write as a single logarithm: 2 ln 4− ln 3.
Solu on
I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42
3
I notln 42
ln 3!
ExamplesExample
Write as a single logarithm: 2 ln 4− ln 3.
Solu on
I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42
3
I notln 42
ln 3!
ExamplesExample
Write as a single logarithm: ln34+ 4 ln 2
Solu on
ln34+ 4 ln 2 = ln 3− ln 4+ 4 ln 2 = ln 3− 2 ln 2+ 4 ln 2
= ln 3+ 2 ln 2 = ln(3 · 22) = ln 12
ExamplesExample
Write as a single logarithm: ln34+ 4 ln 2
Solu on
ln34+ 4 ln 2 = ln 3− ln 4+ 4 ln 2 = ln 3− 2 ln 2+ 4 ln 2
= ln 3+ 2 ln 2 = ln(3 · 22) = ln 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
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
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
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
Change of base formula for logarithmsFact
If 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 aI Therefore
y = loga x =logb xlogb a
Change of base formula for logarithmsFact
If 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 aI Therefore
y = loga x =logb xlogb a
Example of changing base
Example
Find log2 8 by using log10 only.
Solu on
log2 8 =log10 8log10 2
≈ 0.903090.30103
= 3
Surprised? No, log2 8 = log2 23 = 3 directly.
Example of changing base
Example
Find log2 8 by using log10 only.
Solu on
log2 8 =log10 8log10 2
≈ 0.903090.30103
= 3
Surprised? No, log2 8 = log2 23 = 3 directly.
Example of changing base
Example
Find log2 8 by using log10 only.
Solu on
log2 8 =log10 8log10 2
≈ 0.903090.30103
= 3
Surprised?
No, log2 8 = log2 23 = 3 directly.
Example of changing base
Example
Find log2 8 by using log10 only.
Solu on
log2 8 =log10 8log10 2
≈ 0.903090.30103
= 3
Surprised? No, log2 8 = log2 23 = 3 directly.
Upshot of changing baseThe point of the change of base formula
loga x =logb xlogb a
=1
logb a· logb x = constant · logb x
is that all the logarithmic func ons are mul ples of each other. Sojust pick one and call it your favorite.
I Engineers like the common logarithm log = log10I Computer scien sts like the binary logarithm lg = log2I Mathema cians like natural logarithm ln = loge
Naturally, we will follow the mathema cians. Just don’t pronounceit “lawn.”
Summary
I Exponen als turn sums into productsI Logarithms turn products into sumsI Slide rule scabbards are wicked cool
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