﻿ Logistic Model Worksheet | Problems & Solutions Logistic Model Worksheet

Logistic Model Worksheet
• Page 1
1.
Choose the type of function represented by P($t$) = (7.3) · (3.07)$t$. a. Exponential growth function b. Neither exponential growth nor decay function c. Exponential decay function d. None of the above

Solution:

P(t) = (7.3) · (3.07)t is in the form of the exponential function f(t) = x · yt where x = 7.3, y = 3.07

The given function is an exponential growth function.
[ x = 7.3 > 0, and y = 3.07 > 1.]

2.
Choose the type of the function represented by the function, S($t$) = (5.7) · (0.76)$t$. a. Exponential decay function b. Neither exponential growth nor decay function c. Exponential growth function d. None of the above

Solution:

P(t) = (5.7) · (0.76)t is in the form of the exponential function f(t) = a · bt where a = 5.7, b = 0.76

The given function is an exponential decay function.
[ a = 5.7 > 0, and b = 0.76 < 1.]

3.
Which of the following represents the exponential growth function? a. $g$($t$) = (12 )(0.09)$t$ b. $f$($t$) = (1944)(0.09748)$t$ c. $f$($t$) = (7009)(0.994 )$t$ d. $p$($t$) = (3.97)(1.048)$t$

Solution:

The function P(t) = a · btis an exponential growth function if a > 0, b > 1, and is an exponential decay function if a > 0, b < 1.

The function g(t) = (12) · (0.09)t is an exponential decay function.
[a = 12 > 0, b = 0.09 < 1.]

The function f(t) = (1944) · (0.09748)t is an exponential decay function.
[a = 1944 > 0, b = 0.09748 < 1.]

The function f(t) = (7009) · (0.994)t is an exponential decay function.
[a = 7009 > 0, b = 0.994 < 1.]

The function p(t) = (3.97) · (1.048)t is an exponential growth function.
[a = 3.97 > 0, b = 1.048 > 1.]

4.
Which of the following functions represents the exponential decay function? a. $p$($t$) = (2804) · (2.02432)$t$ b. $g$($t$) = (25) · (2.02)$t$ c. $f$($t$) = (2.42) · (0.024)$t$ d. $h$($t$) = (2004) · (1.924)$t$

Solution:

The function P(t) = a · btis an exponential growth function if a > 0, b > 1, and is an exponential decay function if a > 0, b < 1.

The function g(t) = (25) · (2.02)t is an exponential growth function.
[a = 25 > 0, b = 2.02 > 1.]

The function p(t) = (2804) · (2.02432)t is an exponential growth function.
[a = 2804 > 0, b = 2.02432 > 1.]

The function h(t) = (2004) · (1.924)t is an exponential growth function.
[a = 2004 > 0, b = 1.924 > 1.]

The function f(t) = (2.42) · (0.024)t is an exponential decay function.
[a = 2.42 > 0, b = 0.024 < 1.]

5.
Find the percentage rate of growth for the given function, $p$($t$) = (3.4)(1.0545)$t$. a. 0.545 b. 54.5 c. 6.45 d. 5.45

Solution:

p(t) = (3.4) (1.0545)t

Growth factor is1 + r = 1.0545
[Compare with S(t) = So(1 + r)t.]

r = 0.0545

So, the percentage rate of growth is 5.45.

6.
Find the percentage rate of decay for the given function, S($t$) = (2.3)(0.0235)$t$. a. 9.765 b. 97.65 c. 0.0235 d. 2.35

Solution:

S(t) = (2.3)(0.0235)t

Growth factor is 1 + r = 0.0235
[Compare with S(t) = So(1 + r)t.]

r = - 0.9765

So, the percenatge rate of decay is 97.65.

7.
Find rate of growth for the exponential function $g$($t$) = (11) · (1.09)$t$. a. 0.19% b. 0.9% c. 9% d. 1.09%

Solution:

g(t) = (11) · (1.09)t

Growth factor is 1 + r = 1.09
[Compare with S(t) = So(1 + r)t.]

r = 0.09

So, the rate of growth is 9%.

8.
Find the rate of decay for the exponential function $h$($t$) = (5037) · (0.957)$t$. a. 4.3% b. 9.57% c. 0.043% d. 19.57%

Solution:

h(t) = (5037) · (0.957)t

Decay factor is 1 + r = 0.957
[Compare with S(t) = So(1 + r)t.]

r = - 0.043

So, the rate of decay is 4.3%.

9.
A quantity Q grows exponentially over time $t$. At time $t$ = 2, Q = 16 grams and at time $t$ = 5, Q = 128 grams. How much is Q at time $t$ = 3? a. 8 grams b. 16 grams c. 10 grams d. 2 grams

Solution:

As Q grows exponentially over time t, we have Q(t) = Q(0) ekt , where Q(t) is the quantity after time t, Q(0) is the quantity at time t = 0.

At t = 2, Q (2) = 16 grams and at t = 5, Q (5) = 128 grams.
[Given.]

So, Q(5) / Q(2)= 128 / 16 = 8
[Divide Q(5) by Q(2).]

Q(0)e5kQ(0)e2k = 8
[As Q(t) = Q(0)ekt.]

e3k = 8

ek = 2
[Solve for ek .]

From Q(2) = 16 , Q(0) e2k = 16
[Substitute 2 for t in Q(t).]

Q (0) (16) = 16
[Substitute 2 for ek.]

Q (0) = 1
[Simplify.]

At t = 3, Q (3) = Q (0) e3k
[Substitute 3 for t in Q(t).]

= 1 × (ek)3
[Substitute Q (0) = 1.]

= (2)3
[Substitute 3 for ek .]

= 8 grams
[Simplify.]

10.
A bacterial culture grows at a rate proportional to the number of bacteria present. If the size of the culture becomes 7 times every one hour, then how long does it take for the culture to become 8 times? a. 128 minutes b. 67 minutes c. 13 minutes d. 155 minutes

Solution:

Since the culture grows at a rate proportional to the number of bacteria present , the growth of the culture is exponential.

Let A(t) = Aekt, where A(t) be the culture after time t hours, A be the culture at time t = 0.

A(1) = 7A
[Given.]

Aek(1) = 7A
[Substitute 1 for t in A(t).]

ek = 7
[Simplify.]

Let t be the time to make the culture 8 times.

So, A(t) = 8A

A ekt = 8A
[Substitute Aekt for A(t).]

(ek)t = 8
[Simplify.]

7t = 8
[Substitute 7 for ek.]

t = ln8 / ln7 1.07 Hours 67 minutes
[Solve for t.]