34
Chapter 1 Functions
25.
26.
x
2.3219
x
1.3863
¸¸
27.
28.
x
0.6309
x
1.5850
¸±
29. Let t be the number of years. Solving 500,000 1.0375
1,000,000 graphically, we find that t
18.828. The population
ab
t
œ¸
will reach 1 million in about 19 years.
30. (a) The population is given by P t
6250 1.0275 , where t is the number of years after 1890.
a
b
œ
t
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.
31. (a) A t
6.6
ˆ‰
œ
1
2
t14
Î
(b) Solving A t
1 graphically, we find that t
38. There will be 1 gram remaining after about 38.1145 days.
32. Let t be the number of years. Solving 2300 1.06
4150 graphically, we find that t
10.129. It will take about 10.129
t
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 A 1.0625
2A, which
t
œ
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
t
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 A 1
2A,
²œ
0.0625
12
12t
which is equivalent to 1
2. Solving graphically, we find that t
11.119. It will take about 11.119 years. (If
¸
0.0625
12
12t
the interest is credited at the end of each month, it will take 11 years 2 months.)
35. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve Ae
2A, which is
0.0625t
œ
equivalent to e
2. Solving graphically, we find that t
11.090. It will take about 11.090 years.
0.0625t