Embed Size (px)
Citation preview
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
Assume variables are each representing non-zero real numbers
1. Suppose the initial population for some species is 250. Suppose the continuos growth rate is approximately .10annually. Estimate the population for the species after 5 years. Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 250e(.10)t (Substitute given k and P0 values)
P (5) = 250e(.10)5 (Substitute given t value)
P (5) ≈ 412.1803 (calculator)
2. Suppose the initial population for some species is 280. Suppose the continuos growth rate is approximately .10annually. Estimate the population for the species after 30 years. Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 280e(.10)t (Substitute given k and P0 values)
P (30) = 280e(.10)30 (Substitute given t value)
P (30) ≈ 5623.9503 (calculator)
3. Suppose the initial population for some species is 550. Suppose the continuos growth rate is approximately .20annually. Estimate the population for the species after 25 years. Assume the standard population growth model:
P (t) = P0ekt
pg. 1 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
Solution:
P (t) = P0ekt (given)
P (t) = 550e(.20)t (Substitute given k and P0 values)
P (25) = 550e(.20)25 (Substitute given t value)
P (25) ≈ 81627.2375 (calculator)
4. Suppose the initial population for some species is 750. Suppose the continuos growth rate is approximately .25annually. Estimate the population for the species after 10 years. Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 750e(.25)t (Substitute given k and P0 values)
P (10) = 750e(.25)10 (Substitute given t value)
P (10) ≈ 9136.8705 (calculator)
5. Suppose the initial population for some species is 950. Suppose the continuos growth rate is approximately .5annually. Estimate the population for the species after 5 years. Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 950e(.5)t (Substitute given k and P0 values)
P (5) = 950e(.5)5 (Substitute given t value)
P (5) ≈ 11573.3693 (calculator)
pg. 2 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
6. (estimate time to double population) Suppose the initial population for some species is 250. Suppose the continuosgrowth rate is approximately .10 annually. Estimate the number of years for the species population to reach 500.Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 250e(.10)t (Substitute given k and P0 values)
500 = 250e(.10)t (Substitute given population value, P (t))
500
250= e(.10)t (CLM, BI)
ln
(
500
250
)
= (.10)t (def of logs)
1
.10· ln
(
500
250
)
= t (CLM, BI)
t ≈ 6.9315yrs (calculator)
7. (estimate time to double population) Suppose the initial population for some species is 250. Suppose the continuosgrowth rate is approximately .15 annually. Estimate the number of years for the species population to reach 500.Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 250e(.15)t (Substitute given k and P0 values)
500 = 250e(.15)t (Substitute given population value, P (t))
500
250= e(.15)t (CLM, BI)
ln
(
500
250
)
= (.15)t (def of logs)
1
.15· ln
(
500
250
)
= t (CLM, BI)
t ≈ 4.621yrs (calculator)
pg. 3 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
8. (estimate time to triple population) Suppose the initial population for some species is 250. Suppose the continuosgrowth rate is approximately .15 annually. Estimate the number of years for the species population to reach 750.Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 250e(.15)t (Substitute given k and P0 values)
750 = 250e(.15)t (Substitute given population value, P (t))
750
250= e(.15)t (CLM, BI)
ln
(
750
250
)
= (.15)t (def of logs)
1
.15· ln
(
750
250
)
= t (CLM, BI)
t ≈ 7.3241yrs (calculator)
9. Suppose the initial population for some species is 250. Suppose the continuos growth rate is approximately .15annually. Estimate the number of years for the species population to reach 1500. Assume the standard populationgrowth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 250e(.15)t (Substitute given k and P0 values)
1500 = 250e(.15)t (Substitute given population value, P (t))
1500
250= e(.15)t (CLM, BI)
ln
(
1500
250
)
= (.15)t (def of logs)
1
.15· ln
(
1500
250
)
= t (CLM, BI)
t ≈ 11.9451yrs (calculator)
pg. 4 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
10. Suppose the initial population for some species is 280. Suppose the continuos growth rate is approximately .10annually. Estimate the number of years for the species population to reach 3000. Assume the standard populationgrowth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 280e(.10)t (Substitute given k and P0 values)
3000 = 280e(.10)t (Substitute given population value, P (t))
3000
280= e(.10)t (CLM, BI)
ln
(
3000
280
)
= (.10)t (def of logs)
1
.10· ln
(
3000
280
)
= t (CLM, BI)
t ≈ 23.7158yrs (calculator)
11. Suppose the initial population for some species is 550. Suppose the continuos growth rate is approximately .20annually. Estimate the number of years for the species population to reach 2545. Assume the standard populationgrowth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 550e(.20)t (Substitute given k and P0 values)
2545 = 550e(.20)t (Substitute given population value, P (t))
2545
550= e(.20)t (CLM, BI)
ln
(
2545
550
)
= (.20)t (def of logs)
1
.20· ln
(
2545
550
)
= t (CLM, BI)
t ≈ 7.6598yrs (calculator)
pg. 5 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
12. Suppose the initial population for some species is 750. Suppose the continuos growth rate is approximately .25annually. Estimate the number of years for the species population to reach 10900. Assume the standard populationgrowth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 750e(.25)t (Substitute given k and P0 values)
10900 = 750e(.25)t (Substitute given population value, P (t))
10900
750= e(.25)t (CLM, BI)
ln
(
10900
750
)
= (.25)t (def of logs)
1
.25· ln
(
10900
750
)
= t (CLM, BI)
t ≈ 10.7058yrs (calculator)
13. Suppose the initial population for some species is 950. Suppose the continuos growth rate is approximately .5annually. Estimate the number of years for the species population to reach 100000. Assume the standard populationgrowth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
P (t) = 950e(.5)t (Substitute given k and P0 values)
100000 = 950e(.5)t (Substitute given population value, P (t))
100000
950= e(.5)t (CLM, BI)
ln
(
100000
950
)
= (.5)t (def of logs)
1
.5· ln
(
100000
950
)
= t (CLM, BI)
t ≈ 9.3129yrs (calculator)
pg. 6 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
14. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.10 annually. Estimate the number of years for the amount to be half of A0 (this is called the half-lifefor the isotope). Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.10)t (Substitute k value given)
1
2A0 = A0e
(−.10)t (Substitute given amount, A(t) = half of A0)
1
2= e(−.10)t (CLM, BI)
ln
(
1
2
)
= (−.10)t (def of ln)
1
−.10· ln
(
1
2
)
= t (CLM, BI)
6.9315 yrs. ≈ t (calculator)
15. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.15 annually. Estimate the number of years for the amount to be half of A0 (this is called the half-lifefor the isotope). Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
pg. 7 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
A(t) = A0ekt (given)
A(t) = A0e(−.15)t (Substitute k value given)
1
2A0 = A0e
(−.15)t (Substitute given amount, A(t) = half of A0)
1
2= e(−.15)t (CLM, BI)
ln
(
1
2
)
= (−.15)t (def of ln)
1
−.15· ln
(
1
2
)
= t (CLM, BI)
4.621 yrs. ≈ t (calculator)
16. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.55 annually. Estimate the number of years for the amount to be half of A0 (this is called the half-lifefor the isotope). Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.55)t (Substitute k value given)
1
2A0 = A0e
(−.55)t (Substitute given amount, A(t) = half of A0)
1
2= e(−.55)t (CLM, BI)
ln
(
1
2
)
= (−.55)t (def of ln)
1
−.55· ln
(
1
2
)
= t (CLM, BI)
1.2603 yrs. ≈ t (calculator)
17. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.075 annually. Estimate the number of years for the amount to be half of A0 (this is called thehalf-life for the isotope). Assume the standard radioactive decay model:
A(t) = A0ekt
pg. 8 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.075)t (Substitute k value given)
1
2A0 = A0e
(−.075)t (Substitute given amount, A(t) = half of A0)
1
2= e(−.075)t (CLM, BI)
ln
(
1
2
)
= (−.075)t (def of ln)
1
−.075· ln
(
1
2
)
= t (CLM, BI)
9.242 yrs. ≈ t (calculator)
18. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.03 annually. Estimate the number of years for the amount to be half of A0 (this is called the half-lifefor the isotope). Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.03)t (Substitute k value given)
1
2A0 = A0e
(−.03)t (Substitute given amount, A(t) = half of A0)
1
2= e(−.03)t (CLM, BI)
ln
(
1
2
)
= (−.03)t (def of ln)
1
−.03· ln
(
1
2
)
= t (CLM, BI)
23.1049 yrs. ≈ t (calculator)
19. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.02 annually. Estimate the number of years for the amount to be half of A0 (this is called the half-life
pg. 9 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
for the isotope). Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.02)t (Substitute k value given)
1
2A0 = A0e
(−.02)t (Substitute given amount, A(t) = half of A0)
1
2= e(−.02)t (CLM, BI)
ln
(
1
2
)
= (−.02)t (def of ln)
1
−.02· ln
(
1
2
)
= t (CLM, BI)
34.6574 yrs. ≈ t (calculator)
20. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.10 annually. Estimate the number of years for the amount to be 75% of the initial amount, A0.Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.10)t (Substitute k value given)
75
100A0 = A0e
(−.10)t (Substitute given amount, A(t) = 75% of A0)
75
100= e(−.10)t (CLM, BI)
ln
(
75
100
)
= (−.10)t (def of ln)
1
−.10· ln
(
75
100
)
= t (CLM, BI)
2.8768 yrs. ≈ t (calculator)
pg. 10 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
21. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.15 annually. Estimate the number of years for the amount to be 25% of the initial amount, A0.Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.15)t (Substitute k value given)
25
100A0 = A0e
(−.15)t (Substitute given amount, A(t) = 25% of A0)
25
100= e(−.15)t (CLM, BI)
ln
(
25
100
)
= (−.15)t (def of ln)
1
−.15· ln
(
25
100
)
= t (CLM, BI)
9.242 yrs. ≈ t (calculator)
22. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.55 annually. Estimate the number of years for the amount to be 40% of the initial amount, A0.Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.55)t (Substitute k value given)
40
100A0 = A0e
(−.55)t (Substitute given amount, A(t) = 40% of A0)
40
100= e(−.55)t (CLM, BI)
ln
(
40
100
)
= (−.55)t (def of ln)
1
−.55· ln
(
40
100
)
= t (CLM, BI)
1.666 yrs. ≈ t (calculator)
pg. 11 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
23. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.075 annually. Estimate the number of years for the amount to be 60% of the initial amount, A0.Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.075)t (Substitute k value given)
60
100A0 = A0e
(−.075)t (Substitute given amount, A(t) = 60% of A0)
60
100= e(−.075)t (CLM, BI)
ln
(
60
100
)
= (−.075)t (def of ln)
1
−.075· ln
(
60
100
)
= t (CLM, BI)
6.811 yrs. ≈ t (calculator)
24. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.03 annually. Estimate the number of years for the amount to be 80% of the initial amount, A0.Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.03)t (Substitute k value given)
80
100A0 = A0e
(−.03)t (Substitute given amount, A(t) = 80% of A0)
80
100= e(−.03)t (CLM, BI)
ln
(
80
100
)
= (−.03)t (def of ln)
1
−.03· ln
(
80
100
)
= t (CLM, BI)
7.4381 yrs. ≈ t (calculator)
pg. 12 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
25. Suppose the initial amount [in grams] for some radioactive isotope is A0. Suppose the continuos decay rate isapproximately -.02 annually. Estimate the number of years for the amount to be 3% of the initial amount, A0.Assume the standard radioactive decay model:
A(t) = A0ekt
Solution:
A(t) = A0ekt (given)
A(t) = A0e(−.02)t (Substitute k value given)
3
100A0 = A0e
(−.02)t (Substitute given amount, A(t) = 3% of A0)
3
100= e(−.02)t (CLM, BI)
ln
(
3
100
)
= (−.02)t (def of ln)
1
−.02· ln
(
3
100
)
= t (CLM, BI)
175.3279 yrs. ≈ t (calculator)
26. Suppose the initial principal for some deposit is $1000. Suppose after 2 years, the balance is $1500. Estimate theannual interest rate earned by the account.
Assume the standard money growth model for continuous compounding:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
1500 = 1000ek(2) (Substitute t, P0, and P (t) values given)
1500
1000= ek(2) (CLM, BI)
ln
(
1500
1000
)
= k(2) (def of ln)
1
2· ln
(
1500
1000
)
= k (CLM, BI)
0.2027 ≈ k (calculator)
20.27% ≈ k (approx rate of return)
pg. 13 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
27. Suppose the initial principal for some deposit is $1000. Suppose after 20 years, the balance is $150000. Estimate theannual interest rate earned by the account.
Assume the standard money growth model for continuous compounding:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
150000 = 1000ek(20) (Substitute t, P0, and P (t) values given)
150000
1000= ek(20) (CLM, BI)
ln
(
150000
1000
)
= k(20) (def of ln)
1
20· ln
(
150000
1000
)
= k (CLM, BI)
0.2505 ≈ k (calculator)
25.05% ≈ k (approx rate of return)
28. Suppose the initial principal for some deposit is $1000. Suppose after 10 years, the balance is $150000. Estimate theannual interest rate earned by the account.
Assume the standard money growth model for continuous compounding:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
150000 = 1000ek(10) (Substitute t, P0, and P (t) values given)
150000
1000= ek(10) (CLM, BI)
ln
(
150000
1000
)
= k(10) (def of ln)
1
10· ln
(
150000
1000
)
= k (CLM, BI)
0.5011 ≈ k (calculator)
50.11% ≈ k (approx rate of return)
pg. 14 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
29. Suppose the initial principal for some deposit is $1000. Suppose after 30 years, the balance is $1500000. Estimatethe annual interest rate earned by the account.
Assume the standard money growth model for continuous compounding:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
1500000 = 1000ek(30) (Substitute t, P0, and P (t) values given)
1500000
1000= ek(30) (CLM, BI)
ln
(
1500000
1000
)
= k(30) (def of ln)
1
30· ln
(
1500000
1000
)
= k (CLM, BI)
0.2438 ≈ k (calculator)
24.38% ≈ k (approx rate of return)
30. Suppose the initial principal for some deposit is $500. Suppose after 30 years, the balance is $1500000. Estimate theannual interest rate earned by the account.
Assume the standard money growth model for continuous compounding:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
1500000 = 500ek(30) (Substitute t, P0, and P (t) values given)
1500000
500= ek(30) (CLM, BI)
ln
(
1500000
500
)
= k(30) (def of ln)
1
30· ln
(
1500000
500
)
= k (CLM, BI)
0.2669 ≈ k (calculator)
26.69% ≈ k (approx rate of return)
pg. 15 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
31. Suppose the initial population for some bacteria is 75. Suppose after 2 hours, the population is 150. Estimate thehourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
150 = 75ek(2) (Substitute t, P0, and P (t) values given)
150
75= ek(2) (CLM, BI)
ln
(
150
75
)
= k(2) (def of ln)
1
2· ln
(
150
75
)
= k (CLM, BI)
0.3466 ≈ k (calculator)
32. Suppose the initial population for some bacteria is 75. Suppose after 20 hours, the population is 150. Estimate thehourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
150 = 75ek(20) (Substitute t, P0, and P (t) values given)
150
75= ek(20) (CLM, BI)
ln
(
150
75
)
= k(20) (def of ln)
1
20· ln
(
150
75
)
= k (CLM, BI)
0.0347 ≈ k (calculator)
pg. 16 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
33. Suppose the initial population for some bacteria is 75. Suppose after 10 hours, the population is 1500. Estimate thehourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
1500 = 75ek(10) (Substitute t, P0, and P (t) values given)
1500
75= ek(10) (CLM, BI)
ln
(
1500
75
)
= k(10) (def of ln)
1
10· ln
(
1500
75
)
= k (CLM, BI)
0.2996 ≈ k (calculator)
34. Suppose the initial population for some bacteria is 100. Suppose after 20 hours, the population is 5500. Estimatethe hourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
5500 = 100ek(20) (Substitute t, P0, and P (t) values given)
5500
100= ek(20) (CLM, BI)
ln
(
5500
100
)
= k(20) (def of ln)
1
20· ln
(
5500
100
)
= k (CLM, BI)
0.2004 ≈ k (calculator)
pg. 17 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
35. Suppose the initial population for some bacteria is 5. Suppose after 10 hours, the population is 1500. Estimate thehourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
1500 = 5ek(10) (Substitute t, P0, and P (t) values given)
1500
5= ek(10) (CLM, BI)
ln
(
1500
5
)
= k(10) (def of ln)
1
10· ln
(
1500
5
)
= k (CLM, BI)
0.5704 ≈ k (calculator)
36. Suppose the initial population for some bacteria is 10. Suppose after 7 hours, the population is 5500. Estimate thehourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
5500 = 10ek(7) (Substitute t, P0, and P (t) values given)
5500
10= ek(7) (CLM, BI)
ln
(
5500
10
)
= k(7) (def of ln)
1
7· ln
(
5500
10
)
= k (CLM, BI)
0.9014 ≈ k (calculator)
pg. 18 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
37. Suppose the initial population for some bacteria is 10. Suppose after 2 hours, the population is 5500. Estimate thehourly continuous rate of growth, k.
Assume the standard population growth model:
P (t) = P0ekt
Solution:
P (t) = P0ekt (given)
5500 = 10ek(2) (Substitute t, P0, and P (t) values given)
5500
10= ek(2) (CLM, BI)
ln
(
5500
10
)
= k(2) (def of ln)
1
2· ln
(
5500
10
)
= k (CLM, BI)
3.155 ≈ k (calculator)
38. Suppose the initial population for some bacteria is 45. Suppose after 2 hours, the population is 90. Estimatepopulation after 5 hours.
Assume the standard population growth model:
P (t) = P0ekt
Solution: We first calculate the rate of growth, k:
P (t) = P0ekt (given)
90 = 45ek(2) (Substitute t, P0, and P (t) values given)
90
45= ek(2) (CLM, BI)
ln
(
90
45
)
= k(2) (def of ln)
1
2· ln
(
90
45
)
= k (CLM, BI)
0.34657 ≈ k (calculator)
pg. 19 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
We now have a complete growth function with all variables determined:
P (t) = 45e(0.34657)t
Now, we simply plug in t = 5 hours to determine the sought population P (5):
P (t) = 45e(0.34657)t (see work above)
P (5) = 45e(0.34657)5 (substitute t = 5)
≈ 254.6 (calculator)
39. Suppose the initial population for some bacteria is 45. Suppose after 5 hours, the population is 500. Estimatepopulation after 15 hours.
Assume the standard population growth model:
P (t) = P0ekt
Solution: We first calculate the rate of growth, k:
P (t) = P0ekt (given)
500 = 45ek(5) (Substitute t, P0, and P (t) values given)
500
45= ek(5) (CLM, BI)
ln
(
500
45
)
= k(5) (def of ln)
1
5· ln
(
500
45
)
= k (CLM, BI)
0.48159 ≈ k (calculator)
We now have a complete growth function with all variables determined:
P (t) = 45e(0.48159)t
Now, we simply plug in t = 15 hours to determine the sought population P (15):
P (t) = 45e(0.48159)t (see work above)
P (15) = 45e(0.48159)15 (substitute t = 15)
≈ 61729.2 (calculator)
pg. 20 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
40. Suppose the initial population for some bacteria is 75. Suppose after 3 hours, the population is 300. Estimatepopulation after 25 hours.
Assume the standard population growth model:
P (t) = P0ekt
Solution: We first calculate the rate of growth, k:
P (t) = P0ekt (given)
300 = 75ek(3) (Substitute t, P0, and P (t) values given)
300
75= ek(3) (CLM, BI)
ln
(
300
75
)
= k(3) (def of ln)
1
3· ln
(
300
75
)
= k (CLM, BI)
0.4621 ≈ k (calculator)
We now have a complete growth function with all variables determined:
P (t) = 75e(0.4621)t
Now, we simply plug in t = 25 hours to determine the sought population P (25):
P (t) = 75e(0.4621)t (see work above)
P (25) = 75e(0.4621)25 (substitute t = 25)
≈ 7802760.3 (calculator)
41. Suppose the initial population for some bacteria is 55. Suppose after 7 hours, the population is 90. Estimatepopulation after 10 hours.
Assume the standard population growth model:
P (t) = P0ekt
pg. 21 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
Solution: We first calculate the rate of growth, k:
P (t) = P0ekt (given)
90 = 55ek(7) (Substitute t, P0, and P (t) values given)
90
55= ek(7) (CLM, BI)
ln
(
90
55
)
= k(7) (def of ln)
1
7· ln
(
90
55
)
= k (CLM, BI)
0.07035 ≈ k (calculator)
We now have a complete growth function with all variables determined:
P (t) = 55e(0.07035)t
Now, we simply plug in t = 10 hours to determine the sought population P (10):
P (t) = 55e(0.07035)t (see work above)
P (10) = 55e(0.07035)10 (substitute t = 10)
≈ 111.1 (calculator)
42. Suppose the initial population for some bacteria is 51. Suppose after 3 hours, the population is 102. Estimatepopulation after 6 hours.
Assume the standard population growth model:
P (t) = P0ekt
Solution: We first calculate the rate of growth, k:
pg. 22 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
P (t) = P0ekt (given)
102 = 51ek(3) (Substitute t, P0, and P (t) values given)
102
51= ek(3) (CLM, BI)
ln
(
102
51
)
= k(3) (def of ln)
1
3· ln
(
102
51
)
= k (CLM, BI)
0.23105 ≈ k (calculator)
We now have a complete growth function with all variables determined:
P (t) = 51e(0.23105)t
Now, we simply plug in t = 6 hours to determine the sought population P (6):
P (t) = 51e(0.23105)t (see work above)
P (6) = 51e(0.23105)6 (substitute t = 6)
≈ 204 (calculator)
43. Suppose a bottle of wine, initially at 86◦F, is placed in a refrigerator. Suppose after 3 hours, the wine has cooled to75◦F. Furthermore, suppose the refrigerator remains at a constant temperature of 44◦F. Determine the total numberof hours until the wine reaches a chilled temperature of 55◦F.
Assume the wine follows Newton’s Cooling Law:
T (t) = Ts + (T0 − Ts) ekt
Solution:
We first substitute all given values for the variables at the 3 hour mark. This will help us solve for the continuoushourly cooling rate, k. Note the initial temperature for the wine is given by T0 = 86◦F, the surroundingtemperature is given by the temperature inside the refrigerator, Ts = 44◦F. Furthermore, it is given that whent = 3 hours, T (t) = 75◦F. We substitute all of these values below, and solve for k.
pg. 23 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
T (t) = Ts + (T0 − Ts) ekt (given)
75◦ = 44◦ + (86◦ − 44◦) ek(3) (substitute)
75◦ − 44◦ = (86◦ − 44◦) ek(3) (BI)
31◦ = 42◦ek(3) (BI)
31◦
42◦= ek(3) (CLM, BI)
0.7381 = ek(3) (BI)
ln(0.7381) = k(3) (def of ln)
1
3· ln(0.7381) = k (CLM, BI)
−0.1012 ≈ k (calculator)
We now have a complete cooling function with all variables determined, and the general model, T (t) = Ts +(T0 − Ts) e
kt, applied to this scenario becomes:
T (t) = 44◦ + 42◦e(−0.1012)t
Now, we would like to solve for the total number of hours, t, so that the wine temperature is T (t) = 55◦F, thuswe substitute this into our function, and solve for t.
T (t) = 44◦ + 42◦e(−0.1012)t (see work above)
55◦ = 44◦ + 42◦e(−0.1012)t (substitute)
55◦ − 44◦ = 42◦e(−0.1012)t (CLA, BI)
11◦ = 42◦e(−0.1012)t (BI)
11◦
42◦= e(−0.1012)t (CLM, BI)
0.2619 = e(−0.1012)t (BI)
ln(0.2619) = (−0.1012)t (def of ln)
1
−0.1012· ln(0.2619) = t (CLM, BI)
13.2391 hrs. ≈ t (calculator)
44. Suppose a bottle of wine, initially at 72◦F, is placed in a refrigerator. Suppose after 2 hours, the wine has cooled to65◦F. Furthermore, suppose the refrigerator remains at a constant temperature of 40◦F. Determine the total numberof hours until the wine reaches a chilled temperature of 50◦F.
Assume the wine follows Newton’s Cooling Law:
T (t) = Ts + (T0 − Ts) ekt
pg. 24 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
Solution:
We first substitute all given values for the variables at the 2 hour mark. This will help us solve for the continuoushourly cooling rate, k. Note the initial temperature for the wine is given by T0 = 72◦F, the surroundingtemperature is given by the temperature inside the refrigerator, Ts = 40◦F. Furthermore, it is given that whent = 2 hours, T (t) = 65◦F. We substitute all of these values below, and solve for k.
T (t) = Ts + (T0 − Ts) ekt (given)
65◦ = 40◦ + (72◦ − 40◦) ek(2) (substitute)
65◦ − 40◦ = (72◦ − 40◦) ek(2) (BI)
25◦ = 32◦ek(2) (BI)
25◦
32◦= ek(2) (CLM, BI)
0.7813 = ek(2) (BI)
ln(0.7813) = k(2) (def of ln)
1
2· ln(0.7813) = k (CLM, BI)
−0.1234 ≈ k (calculator)
We now have a complete cooling function with all variables determined, and the general model, T (t) = Ts +(T0 − Ts) e
kt, applied to this scenario becomes:
T (t) = 40◦ + 32◦e(−0.1234)t
Now, we would like to solve for the total number of hours, t, so that the wine temperature is T (t) = 50◦F, thuswe substitute this into our function, and solve for t.
T (t) = 40◦ + 32◦e(−0.1234)t (see work above)
50◦ = 40◦ + 32◦e(−0.1234)t (substitute)
50◦ − 40◦ = 32◦e(−0.1234)t (CLA, BI)
10◦ = 32◦e(−0.1234)t (BI)
10◦
32◦= e(−0.1234)t (CLM, BI)
0.3125 = e(−0.1234)t (BI)
ln(0.3125) = (−0.1234)t (def of ln)
1
−0.1234· ln(0.3125) = t (CLM, BI)
9.4259 hrs. ≈ t (calculator)
45. Suppose a bottle of wine, initially at 72◦F, is placed in a refrigerator. Suppose after 4 hours, the wine has cooled to
pg. 25 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
65◦F. Furthermore, suppose the refrigerator remains at a constant temperature of 40◦F. Determine the total numberof hours until the wine reaches a chilled temperature of 50◦F.
Assume the wine follows Newton’s Cooling Law:
T (t) = Ts + (T0 − Ts) ekt
Solution:
We first substitute all given values for the variables at the 4 hour mark. This will help us solve for the continuoushourly cooling rate, k. Note the initial temperature for the wine is given by T0 = 72◦F, the surroundingtemperature is given by the temperature inside the refrigerator, Ts = 40◦F. Furthermore, it is given that whent = 4 hours, T (t) = 65◦F. We substitute all of these values below, and solve for k.
T (t) = Ts + (T0 − Ts) ekt (given)
65◦ = 40◦ + (72◦ − 40◦) ek(4) (substitute)
65◦ − 40◦ = (72◦ − 40◦) ek(4) (BI)
25◦ = 32◦ek(4) (BI)
25◦
32◦= ek(4) (CLM, BI)
0.7813 = ek(4) (BI)
ln(0.7813) = k(4) (def of ln)
1
4· ln(0.7813) = k (CLM, BI)
−0.0617 ≈ k (calculator)
We now have a complete cooling function with all variables determined, and the general model, T (t) = Ts +(T0 − Ts) e
kt, applied to this scenario becomes:
T (t) = 40◦ + 32◦e(−0.0617)t
Now, we would like to solve for the total number of hours, t, so that the wine temperature is T (t) = 50◦F, thuswe substitute this into our function, and solve for t.
T (t) = 40◦ + 32◦e(−0.0617)t (see work above)
50◦ = 40◦ + 32◦e(−0.0617)t (substitute)
50◦ − 40◦ = 32◦e(−0.0617)t (CLA, BI)
10◦ = 32◦e(−0.0617)t (BI)
10◦
32◦= e(−0.0617)t (CLM, BI)
0.3125 = e(−0.0617)t (BI)
ln(0.3125) = (−0.0617)t (def of ln)
1
−0.0617· ln(0.3125) = t (CLM, BI)
18.8517 hrs. ≈ t (calculator)
pg. 26 c©2007-2011 MathHands.com v.1110
INTERMEDIATE ALGEBRACH 06 SEC 7 SOLUTIONS
mathhands
Log Functions
models
pg. 27 c©2007-2011 MathHands.com v.1110
