Calculus 2nd Edition Rogawski

SECTION 7.3 Logarithms and Their Derivatives 357
Solution
(a) log 6 9 + log 6 4 = log 6 (9 · 4) = log 6 (36) = log 6 (6 2 ) = 2
( ) 1√e
(b) ln = ln(e −1/2 ) = − 1 2 ln(e) = −1 2
(c) 10 log b (b 3 ) − 4 log b ( √ ( 1
b) = 10(3) − 4 log b (b 1/2 ) = 30 − 4 = 28
2)
EXAMPLE 2 Solving an Exponential Equation The bacteria population in a bottle at
time t (in hours) has size P (t) = 1000e 0.35t . After how many hours will there be 5000
bacteria?
Bacteria population P
6000
5000
4000
3000
2000
1000
t (h)
1 2 3 4 4.6
FIGURE 3 Bacteria population as a function
of time.
Solution We must solve P (t) = 1000e 0.35t = 5000 for t (Figure 3):
e 0.35t = 5000
1000 = 5
ln(e 0.35t ) = ln 5
(take logarithms of both sides)
0.35t = ln 5 ≈ 1.609 [because ln(e a ) = a]
t ≈ 1.609
0.35
≈ 4.6 hours
Calculus of Logarithms
In Section 7.1, we proved that for any base b>0,
d
dx bx = m(b) b x b h − 1
, where m(b) = lim
h→0 h
REMINDER ln x is the natural
logarithm, that is, ln x = log e x.
However, we were not able to identify the factor m(b) (other than to say that e is the unique
number for which m(e) = 1). Now we can use the Chain Rule to prove that m(b) = ln b.
The key point is that every exponential function can be written in terms of e, namely,
b x = (e ln(b) ) x = e (ln b)x . By the Chain Rule,
d
dx bx = d
dx e(ln b)x = (ln b)e (ln b)x = (ln b)b x
THEOREM 1 Derivative of f (x) = b x
d
dx bx = (ln b)b x for b>0 2
For example, (10 x ) ′ = (ln 10)10 x .
EXAMPLE 3 Differentiate: (a) f (x) = 4 3x and (b) f (x) = 5 x2 .
Solution
(a) The function f (x) = 4 3x is a composite of 4 u and u = 3x:
d
dx 43x =
( d
du 4u ) du
dx = (ln 4)4u (3x) ′ = (ln 4)4 3x (3) = (3ln4)4 3x