Notesale: Turn your study into money

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Exponentials and Logarithms (naturally)
In this chapter we first recall some facts about exponentials (xy with x > 0 and y arbitrary): they should
y
be familiar from algebra, or “precalculus
...
g
...
718 281 828 459 045 235 360 287 471 352 662 497 757 247 093 699 95 · · ·
...
”
1
...

For any real number x and any positive integer n = 1, 2, 3,
...

To define xp/q for a general fraction
(37)

p

q

1


...
One then defines
√
xp/q = q xp
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One can define xa for irrational
numbers a by taking limits
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e
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4 = 14
,
a3 = 1
...

√a4 = 1
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g
...
Our definition of 2 2 then is
√

2
i
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we define 2

√

2

= lim 2a n ,
n→∞

2

as the limit of the sequence of numbers
√
√
√
10
100
1000
2, 214 ,
2141 ,
21414 , · · ·

(See table 1
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We will not go into these details in this course
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One can show that these properties still hold if a and b are real numbers (not
necessarily fractions
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85

1
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4000000000
1
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4140000000
1
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4142100000
1
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4142135000

2
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639015821546
2
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664749650184
2
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665137561794

√
√
Table 1
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Note that as x gets
closer to 2 the quantity 2x appears to converge to
√
some number
...


Now instead of considering xa as a function of x we can pick a positive number a and consider the
function f (x) = a x
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1
...
The trouble with powers of negative numbers
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For instance 3 —8 = —
2 because ( —2)3 = —8
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But there is a problem: since 62 = 31 you would think that (—8)2/ 6 = (—8)1/3
...

Another example:
√
(—4)1/2 = —4 is not defined
but, even though 1 = 2 ,
2
4
√
√
(—4)2/ 4 = 4 (—4)2 = 4 +16 = 2 is defined
...

The safest is just not to take fractional powers of negative numbers
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For example, (—8)π is not defined1
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Logarithms
Briefly, y = loga x is the inverse function to y = ax
...

In other words, loga x is the answer to the question “for which number y does one have x = ay?” The number
loga x is called the logarithm with base a of x
...

For instance,
√
1
23 = 8, 21/ 2 = 2, 2−1 =
2
so
√
1
1
log2 8 = 3, log2 2 = , log2 = —1
...
You will see this next semester if you take math 222
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The graphs of y = 2x, 3x, (1/2)x, (0
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The graphs are purposely not
labeled: can you figure out which is which?

Also:
log2(—3) doesn’t exist
because there is no number y for which 2y = —3 (2y is always positive) and
log−3 2 doesn’t exist either
because y = log−3 2 would have to be some real number which satisfies (—3)y = 2, and we don’t take
non-integer powers of negative numbers
...
Properties of logarithms
In general one has
loga ax = x, and aloga x = x
...

Again, one finds the following formulas in precalculus texts:

loga
(39)

y

loga x — loga y

b

They follow from (38)
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Graphs of exponential functions and logarithms
Figure 1 shows the graphs of some exponential functions y = ax with different values of a, and figure 2
shows the graphs of y = log2 x, y = log3 x, log1/2 x, log1/3 (x) and y = log10 x
...
)
From algebra/precalc recall:

and

In other words, for a > 1 it follows from x1 < x2 that ax1 < ax2 ; if 0 < a < 1, then x1 < x2 implies ax1 > ax2
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Graphs of some logarithms
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Can you tell what a is for each graph?

5
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∆x→0
∆x
So if we assume that the limit
2∆x — 1 = C
∆x→0
∆x
lim

exists then we have
d2x
= C2x
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693 147
...


Once we know (40) we can compute the derivative of ax for any other positive number a
...


By the chain rule we therefore get
dax = d2x·log2 a
dx
dx

dx · log2 a
dx
= (C log2 a) 2 x·log2 a
= (C log2 a) ax
...
This is essentially our
formula for the derivative of ax, but one can make the formula look nicer by introducing a special number,
namely, we define
2∆x — 1

...
718 281 818 459 · · ·
This number is special because if you set a = e, then
1
= 1,
C log2 a = C log2 e = C log2 21/C = C ·
C
and therefore the derivative of the function y = ex is
(41)
Read that again: the function ex is its own derivative!
The logarithm with base e is called the Natural Logarithm, and is written
ln x = loge x
...

For any positive number a we have a = eln a, and also
ax = ex ln a
...
Derivatives of Logarithms
Since the natural logarithm is the inverse function of f (x) = ex we can find its derivative by implicit
differentiation
...

Then solve for f ′(x) to get

1

f ′(x) =

Finally we remember that af(x)


...

dx
x ln a
89

In particular, the natural logarithm has a very simple derivative, namely, since ln e = 1 we have
(44)

7
...
1
...
Let r be any real number
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e
...

x→∞ a
This theorem says that any exponential will beat any power of x as x → ∞
...
001)x go to infinity, but
1000
= 0,
lim x
x
x→∞ (1
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001x will be much larger than 1000x
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We want to show lim x→∞ xr e− x = 0
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f ′(x) =

dxr+1e−x
= (r + 1)xr — xr+1 e− x = (r + 1 — x)xre−x
...
e
...
It follows that f (x) < f (r + 1) for all
x > r + 1, i
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xr+1e−x < (r + 1)r+1e−(r+1) for x > r + 1
...


x
The Sandwich Theorem implies that limx→∞ xre− x = 0, which is what we had promised to show
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N
...
)
x→∞ x
ln x
m > 0 = ⇒ lim
=0
x→∞ xm
m > 0 =⇒ lim xm ln x = 0
x→0

The second limit says that even though ln x becomes infinitely large as x → ∞ , it is always much less than
any power xm with m > 0 real
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m t→∞ es

The third limit follows from the second by substituting x = 1/y and using ln
90

1

x

= — ln x
...
Exponential growth and decay
A quantity X which depends on time t is said to grow or decay exponentially if it is given by
X(t) = X0ekt
...

The derivative of an exponentially growing quantity, i
...
its rate of change with time, is given by
X′(t) = X0 kekt so that
dX(t)

= kX(t)
...

The proportionality constant is k and is sometimes called “the relative growth rate
...
To see that this is true, suppose you have a function X(t) for
which X′(t) = kX(t) holds at all times t
...

It follows that X(t)e− kt does not depend on t
...

Multiply with ekt and we end up with
X(t) = X0ekt
...
1
...
If X(t) = X0ekt then one has
X(t + T ) = X0 ekt+kT = X0 ektekT = ekT X(t)
...
If
k > 0, so that the quantity is actually growing, then one calls
T =

ln 2
k

the doubling time for X because X(t) changes by a factor ekT = eln 2 = 2 every T time units: X(t) doubles
every T time units
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2

8
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Determining X0 and k
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Suppose that you know
X1 = X(t1) and X2 = X(t2)
...
One first finds k from
X1
X0ekt1
X
=
= ek(t1−t2) = ⇒ ln 1 = k(t — 1 — t2)
kt
X2
X2
X0 e 2
which implies
ln X1 — ln X2
k=

...

0
ekt1
ekt2
(both expressions should give the same result
...
Exercises

r

298
...


1+x
1—x

(Hint for some of these: if you have to solve something like e4x — 3e3x + ex = 0, then call w = ex, and you
get a polynomial equation for w, namely w4 — 3w3 + w =
0
...
y = ln 1 + x2
300
...
y = ex

301
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y = e−x

302
...
Sketch the graph of f
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y = ex + e−2x
285
...
Sketch the part of the graph of the function

ex
286
...
y =
1 + e2x

1

f (x) = e− x
with x > 0
...
y = xe−x
√ −x/4
289
...
?)

290
...
A damped oscillation is a function of the form

291
...
y = ln x
1
293
...
y = x ln x
—1
295
...

Sketch the graph of f (x) = e−x sin 10x (i
...
find zeroes, local max and mins, inflection points) and draw (with
pencil on paper) the piece of the graph with 0 ≤ x ≤ 2π
...

What is the ratio between the function values at two
consecutive local maxima? (Hint: the answer does not
depend on which pair of consecutive local maxima you
consider
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y = (ln x)2 (x > 0)
ln x
(x > 0)
297
...
Find the inflection points on the graph of f (x) =
(1 + x) ln x (x > 0)
...
Polonium-210 has a half life of 140 days
...

(b) Find the mass after 100 days
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306
...
How many more intersections do
these graphs have (with —∞ < x < ∞)?

307
...

309
...

311
...

313
...


325
...
The 1950 world
population was 2
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4 billion
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ex — 1
lim x
x→∞ e + 1
ex — x2
lim
x→∞ ex + x
2x
lim
x→∞ 3 x — 2x
ex — x2
lim
x→∞ e2x + e−x
lim

x→∞

lim

−x

x→∞

The ACME company runs two ads on Sunday mornings
...
” What does ACME
think the population of the world is at present? How
fast does ACME think the population is increasing? Use
units of billions of people so you can write 8 instead of
8, 000, 000, 000
...
)

−x/2

e √—e
ex + 1
√
x + e4x
e2x + x

x→∞

lim

326
...


√

e

√

x

327
...
The population of
California on January 1, 2000 was 20,000,000
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lim ln x
x→∞ ln x2
316
...


(b) Each Californian consumes pizzas at the rate of
70 pizzas per year
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lim √

x→∞

318
...
The population of the country of Farfarawa¨y grows
exponentially
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Find the tenth derivative of xex
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For which real number x is 2x — 3x the largest?
321
...

, and
dx
dx
dx

(b) How long will it take the population to double?
(Your answer may be expressed in terms of exponentials
and natural logarithms
...


322
...

About logarithmic differentiation:
(a) Let y = (x + 1)2(x + 3)4(x + 5)6 and u = ln y
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Hint: Use the fact that ln converts multiplication
to addition before you differentiate
...

(b) Check that the derivative of ln u(x) is the logarithmic
derivative of the function u (as defined in the exercises
following §25, chapter 4
...
The hyperbolic functions are defined by
ex — e−x ,
2
ex + e−x ,
cosh x =
2
sinh x
tanh x =

...
After 3 days a sample of radon-222 decayed to 58%
of its original amount
...

93

(b) Show that

(c) Sketch the graphs of the three hyperbolic functions
...

dx
cosh2 x

94

95



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