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With this intuitive notion we readily see that
log2 16 = “the power to which we raise 2 to obtain 16” = 4
and
log2 (1/4) = “the power to which we raise 2 to obtain 1/4” = ’2.
However this intuitive approach does not work so well if we want to take logπ 5

or log2 7. Therefore we will give a new de¬nition of the logarithm to any base
a > 0 which in simple cases coincides with the intuitive notion of logarithm.
If a > 0 and b > 0 then
ln b
loga b = .
ln a
EXAMPLE 6.19
Calculate log2 32.

SOLUTION
We see that
5 · ln 2
ln 25
ln 32
log2 32 = = = = 5.
ln 2 ln 2 ln 2
Notice that, in this example, the new de¬nition of log2 32 agrees with the
intuitive notion just discussed.
EXAMPLE 6.20
Express ln x as the logarithm to some base.

SOLUTION
If x > 0 then
ln x ln x
loge x = = = ln x.
ln e 1
Thus we see that the natural logarithm ln x is precisely the same as loge x.

Math Note: In mathematics, it is common to write ln x rather than loge x.

You Try It: Calculate log3 27 + log5 (1/25) ’ log2 8.
164 CHAPTER 6 Transcendental Functions

We will be able to do calculations much more easily if we learn some simple
properties of logarithms and exponentials.
If a > 0 and b > 0 then
a (loga b) = b.
If a > 0 and b ∈ R is arbitrary then
loga (a b ) = b.
If a > 0, b > 0, and c > 0 then
(i) loga (b · c) = loga b + loga c
(ii) loga (b/c) = loga b ’ loga c
logc b
(iii) loga b =
logc a
1
(iv) loga b =
logb a
(v) loga 1 = 0
(vi) loga a = 1
(vii) For any exponent ±, loga (b± ) = ± · (loga b)
We next give several examples to familiarize you with logarithmic and
exponential operations.

EXAMPLE 6.21
Simplify the expression

log3 81 ’ 5 · log2 8 ’ 3 · ln(e4 ).

SOLUTION
The expression equals

log3 (34 )’5·log2 (23 )’3·ln e4 = 4·log3 3’5·[3·log2 2]’3·[4·ln e]
= 4·1’5·3·1’3·4·1 = ’23.

You Try It: What does log3 5 mean in terms of natural logarithms?

EXAMPLE 6.22
Solve the equation
4
5x · 23x =
7x
for the unknown x.
165
CHAPTER 6 Transcendental Functions

SOLUTION
We take the natural logarithm of both sides:
4
ln(5x · 23x ) = ln .
7x
Applying the rules for logarithms we obtain
ln(5x ) + ln(23x ) = ln 4 ’ ln(7x )
or
x · ln 5 + 3x · ln 2 = ln 4 ’ x · ln 7.
Gathering together all the terms involving x yields
x · [ln 5 + 3 · ln 2 + ln 7] = ln 4
or
x · [ln(5 · 23 · 7)] = ln 4.
Solving for x gives
ln 4
x= = log280 4.
ln 280
EXAMPLE 6.23
Simplify the expression
5 · log7 3 ’ (1/4) · log7 16
B= .
3 · log7 5 + (1/5) · log7 32

SOLUTION
The numerator of B equals

log7 (35 ) ’ log7 (161/4 ) = log7 243 ’ log7 2 = log7 (243/2).
Similarly, the denominator can be rewritten as
log7 53 + log7 (321/5 ) = log7 125 + log7 2 = log7 (125 · 2) = log7 250.
Putting these two results together, we ¬nd that
log7 243/2
B= = log250 (243/2).
log7 250

32
You Try It: What does mean (in terms of the natural logarithm function)?
EXAMPLE 6.24
Simplify the expression (log4 9) · (log9 16).
166 CHAPTER 6 Transcendental Functions

SOLUTION
We have
1
(log4 9) · (log9 15) = · log9 16 = log4 16 = 2.
log9 4



6.4 Calculus with Logs and Exponentials to
Arbitrary Bases
6.4.1 DIFFERENTIATION AND INTEGRATION OF
loga x AND ax
We begin by noting these facts:
If a > 0 then
ax
dx
a = a x · ln a; equivalently, a dx = + C.
x
(i)
ln a
dx
1
d
(loga x) =
(ii)
x · ln a
dx
Math Note: As always, we can state these last formulas more generally as
du du
a = au · · ln a
dx dx
and
1 du 1
d
loga u = · · .
u dx ln a
dx
EXAMPLE 6.25
Calculate
d d d d
(log4 (x · cos x)).
(3cos x ),
(5x ), (log8 x),
dx dx dx dx
SOLUTION
We see that
dx
(5 ) = 5x · ln 5.
dx
For the second problem, we apply our general formulation with a = 3, u =
cos x to obtain
d cos x d
) = 3cos x · cos x · ln 3 = 3cos x · (’ sin x) · ln 3.
(3
dx dx
167
CHAPTER 6 Transcendental Functions

Similarly,
1
d
(log8 x) =
x · ln 8
dx
1
d d
(log4 (x · cos x)) = · (x · cos x)
(x · cos x) · ln 4 dx
dx
cos x + (x · (’ sin x))
= .
(x · cos x) · ln 4
EXAMPLE 6.26
Integrate

3cot x · (’ csc2 x) dx.

SOLUTION
For clarity we set •(x) = cot x, • (x) = ’ csc2 x. Then our integral becomes
1
3•(x) · • (x) dx = · 3•(x) · • (x) · ln 3 dx
ln 3
1
= · 3•(x) + C.
ln 3
Resubstituting the expression for •(x) now gives that
1
3cot x · (’ csc2 x) dx = · 3cot x + C.
ln 3

You Try It: Evaluate (log6 (x 3 )/x) dx.

You Try It: Calculate the integral
2
x · 3x dx.

Our new ideas about arbitrary exponents and bases now allow us to formulate a
general result about derivatives of powers:
For any real exponent a we have
da
x = a · x a’1 .
dx
EXAMPLE 6.27 √
’π, x 3, x e.
Calculate the derivative of x
168 CHAPTER 6 Transcendental Functions

SOLUTION
We have
d ’π
= ’π · x ’π ’1 ,
x
dx
d √3 √ √
x = 3 · x 3’1 ,
dx
de
x = e · x e’1 .
dx
2
You Try It: Calculate (d/dx)5sin x’x . Calculate (d/dx)x 4π .

6.4.2 GRAPHING OF LOGARITHMIC AND
EXPONENTIAL FUNCTIONS
If a > 0 and f (x) = loga x, x > 0, then



Y
1
f (x) =
x · ln a
FL
’1
f (x) = 2
x · ln a
AM

f (1) = 0.
Using this information, we can sketch the graph of f (x) = loga x.
If a > 1 then ln a > 0 so that f (x) > 0 and f (x) < 0. The graph of f is
TE



exhibited in Fig. 6.6.




Fig. 6.6

If 0 < a < 1 then ln a = ’ ln(1/a) < 0 so that f (x) < 0 and f (x) > 0. The
graph of f is sketched in Fig. 6.7.
Since g(x) = a x is the inverse function to f (x) = loga x, the graph of g is the
re¬‚ection in the line y = x of the graph of f (Figs 6.6 and 6.7). See Figs 6.8, 6.9.
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CHAPTER 6 Transcendental Functions




Fig. 6.7




Fig. 6.8




Fig. 6.9


Figure 6.10 shows the graphs of loga x for several different values of a > 1.
Figure 6.11 shows the graphs of a x for several different values of a > 1.

You Try It: Sketch the graph of y = 4x and y = log4 x on the same set of axes.
170 CHAPTER 6 Transcendental Functions




Fig. 6.10




Fig. 6.11

6.4.3 LOGARITHMIC DIFFERENTIATION
We next show how to use the logarithm as an aid to differentiation. The key idea is
that if F is a function taking positive values then we can exploit the formula
F
[ln F ] = (—)
.
F
EXAMPLE 6.28
Calculate the derivative of the function
F (x) = (cos x)(sin x) , 0 < x < π.
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CHAPTER 6 Transcendental Functions

SOLUTION
We take the natural logarithm of both sides:

ln F (x) = ln (cos x)(sin x) = (sin x) · (ln(cos x)). (†)
Now we calculate the derivative using the formula (—) preceding this example:
The derivative of the left side of (†) is
F (x)
.
F (x)
Using the product rule, we see that the derivative of the far right side of (†) is
’ sin x
(cos x) · (ln(cos x)) + (sin x) · .
cos x
We conclude that
’ sin x
F (x)
= (cos x) · (ln(cos x)) + (sin x) · .
cos x
F (x)
Thus
sin2 x
F (x) = (cos x) · (ln(cos x)) ’ · F (x)
cos x
sin2 x
= (cos x) · ln(cos x) ’ · (cos x)(sin x)
cos x

You Try It: Differentiate log9 | cos x|.

You Try It: Differentiate 3sin(3x) . Differentiate x sin 3x .
EXAMPLE 6.29
Calculate the derivative of F (x) = x 2 · (sin x) · 5x .

SOLUTION
We have

[ln F (x)] = [ln(x 2 · (sin x) · 5x )]
= [(2 · ln x) + ln(sin x) + (x · ln 5)]
2 cos x
=+ + ln 5.
sin x
x
Using formula (—), we conclude that
2 cos x
F (x)
=+ + ln 5
sin x
F (x) x
172 CHAPTER 6 Transcendental Functions

hence
2 cos x
F (x) = + + ln 5 · [x 2 · (sin x) · 5x ].
sin x
x

You Try It: Calculate (d/dx)[(ln x)ln x ].




6.5 Exponential Growth and Decay
Many processes of nature and many mathematical applications involve logarithmic
and exponential functions. For example, if we examine a population of bacteria,
we notice that the rate at which the population grows is proportional to the number
of bacteria present. To see that this makes good sense, suppose that a bacterium
reproduces itself every 4 hours. If we begin with 5 thousand bacteria then
after 4 hours there are 10 thousand bacteria
after 8 hours there are 20 thousand bacteria
after 12 hours there are 40 thousand bacteria
after 16 hours there are 80 thousand bacteria . . .
etc.
The point is that each new generation of bacteria also reproduces, and the older
generations reproduce as well.Asketch (Fig. 6.12) of the bacteria population against
time shows that the growth is certainly not linear”indeed the shape of the curve
appears to be of exponential form.




Fig. 6.12
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CHAPTER 6 Transcendental Functions

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