Thomas Calculus 13th [Solutions]

Section 1.5 Exponential Functions 33
25. 26.
x 2.3219 x 1.3863
27. 28.
x 0.6309 x 1.5850
t
29. Let t be the number of years. Solving 500,000(1.0375) 1,000,000
population will reach 1 million in about 19 years.
graphically, we find that t 18.828. The
t
30. (a) The population is given by P( t ) 6250(1.0275) , where t is the number of years after 1890.
Population in 1915: P(25) 12,315
Population in 1940: P(50) 24,265
(b) Solving P(t) = 50,000 graphically, we find that t 76.651. The population reached 50,000 about 77 years
after 1890, in 1967.
/14
31. (a)
1
t
A( t) 6.6
2
(b) Solving A(t) = 1 graphically, we find that t 38. There will be 1 gram remaining after about 38.1145 days.
t
32. Let t be the number of years. Solving 2300(1.60) 4150 graphically, we find that t 10.129. It will take
about 10.129 years. (If the interest is not credited to the account until the end of each year, it will take
11 years.)
33. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve
t
t
A(1.0625) 2 A , which is equivalent to 1.0625 2. Solving graphically, we find that t 11.433. It will take
about 11.433 years. (If the interest is credited at the end of each year, it will take 12 years.)
34. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve
0.0575t
0.0575t
Ae 3 A , which is equivalent to e 3. Solving graphically, we find that t 19.106. It will take
about 19.106 years.
Copyright
2014 Pearson Education, Inc.