Solução_Calculo_Stewart_6e

F.
TX.10
30 ¤ CHAPTER 1 FUNCTIONS AND MODELS
31. (a) It is defined as the inverse of the exponential function with base a,thatis,log a x = y ⇔ a y = x.
(b) (0, ∞) (c) R (d) See Figure 11.
33. (a) log 5 125 = 3 since 5 3 =125. (b) log 3
1
27 = −3 since 3−3 = 1 3 3 = 1 27 .
35. (a) log 2 6 − log 2 15 + log 2 20 = log 2 ( 6
15 )+log 2 20 [by Law 2]
=log 2 ( 6 · 20) [by Law 1]
15
=log 2 8,and log 2 8=3since 2 3 =8.
(b) log 3 100 − log 3 18 − log 3 50 = log 100
3 18 − log3 50 = log 100
3 18·50
37. ln5+5ln3=ln5+ln3 5 [by Law 3]
=ln(5· 3 5 ) [by Law 1]
=ln1215
=log 3 ( 1 9 ),and log 3 1
9
= −2 since 3 −2 = 1 9 .
39. ln(1 + x 2 )+ 1 2 ln x − ln sin x =ln(1+x2 )+lnx 1/2 − ln sin x =ln[(1+x 2 ) √ x ] − ln sin x =ln (1 + x2 ) √ x
sin x
41. To graph these functions, we use log 1.5 x = ln x
ln 1.5 and log 50 x = ln x
ln 50 .
These graphs all approach −∞ as x → 0 + , and they all pass through the
point (1, 0). Also, they are all increasing, and all approach ∞ as x →∞.
The functions with larger bases increase extremely slowly, and the ones with
smaller bases do so somewhat more quickly. The functions with large bases
approach the y-axis more closely as x → 0 + .
43. 3 ft =36in, so we need x such that log 2 x =36 ⇔ x =2 36 =68,719,476,736. Inmiles,thisis
68,719,476,736 in · 1 ft
12 in · 1 mi
≈ 1,084,587.7 mi.
5280 ft
45. (a) Shift the graph of y =log 10 x five units to the left to
obtain the graph of y =log 10 (x +5).Notethevertical
asymptote of x = −5.
(b) Reflect the graph of y =lnx about the x-axis to obtain
the graph of y = − ln x.
y =log 10 x y =log 10 (x +5)
47. (a) 2lnx =1 ⇒ ln x = 1 2
⇒ x = e 1/2 = √ e
(b) e −x =5 ⇒ −x =ln5 ⇒ x = − ln 5
y =lnx
y = − ln x
49. (a) 2 x−5 =3 ⇔ log 2 3=x − 5 ⇔ x =5+log 2 3.
Or: 2 x−5 =3 ⇔ ln 2 x−5 =ln3 ⇔ (x − 5) ln 2 = ln 3 ⇔ x − 5= ln 3
ln 2
⇔ x =5+ ln 3
ln 2