﻿ Exponential Decay Functions Worksheet | Problems & Solutions
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Exponential Decay Functions Worksheet
• Page 1
1.
State whether the given model is an exponential growth model or an exponential decay model.
$v$ = 0.12(4)$t$ a. Exponential Growth Model b. Exponential Decay Model c. Both d. None of the above

#### Solution:

Substitute 0, 1, 2, and 3 for t in the exponential function v = 0.12(4)t

v = 0.12(4)0
[Substitute 0 for t.]

= 0.12 x 1 = 0.12
[40 = 1.]

v = 0.12(4)1
[Substitute 1 for t.]

= 0.12 x 4 = 0.48
[41 = 4.]

v = 0.12(4)2
[Substitute 2 for t.]

= 0.12 x 16 = 1.92
[42 = 16.]

v = 0.12(4)3
[Substitute 3 for t.]

= 0.12 x 64 = 7.68
[43 = 64.]

The value of v increases by the same percent in each unit of time.

The model v = 0.12(4)t is an exponential growth model.

Correct answer : (1)
2.
State whether the model $y$ = 2(0.12)$t$ is an exponential growth model or an exponential decay model. a. Exponential Growth Model b. Exponential Decay Model c. Both d. None of the above

#### Solution:

Substitute 0, 1, 2, and 3 for t in the exponential function y = 2(0.12)t.

y = 2(0.12)0
[Substitute 0 for t.]

= 2 x 1 = 2
[(0.12)0 = 1.]

y = 2(0.12)1
[Substitute 1 for t.]

= 2 x 0.12 = 0.24
[(0.12)1 = (0.12)]

y = 2(0.12)2
[Substitute 2 for t.]

= 2 x 0.12 x 0.12 = 0.028
[Expand (0.12)2.]

y = 2(0.12)3
[Substitute 3 for t.]

= 2 x 0.12 x 0.12 x 0.12 = 0.002
[Expand (0.12)3.]

The value of y decreases by the same percent in each unit of time.

The model y = 2(0.12)t is an exponential decay model.

Correct answer : (2)
3.
State whether the model $y$ = 13(0.2)$t$ is an exponential growth model or an exponential decay model. a. Exponential growth model b. Exponential decay model c. Both

#### Solution:

Substitute 0, 1, 2, and 3 for t in the exponential function y = 13(0.2)t.

y = 13(0.2)0
[Substitute 0 for t.]

= 13 × 1 = 13
[(0.2)0 = 1]

y = 13(0.2)1
[Substitute 1 for t.]

= 13 × 0.2 = 2.6
[(0.2)1 = 0.2]

y = 13(0.2)2
[Substitute 2 for t.]

= 13 × 0.2 × 0.2 = 0.520
[Expand (0.2)2.]

y = 13(0.2)3
[Substitute 3 for t.]

= 13 × 0.2 × 0.2 × 0.2 = 0.104
[Expand (0.2)3.]

The value of y decreases by the same percent of 0.2 in each unit of time.

A quantity displays exponential decay, if it decreases by the same percent r in each unit of tme t.

So, the model y = 13(0.2)t is an exponential decay model.

Correct answer : (2)
4.
State whether the model $y$ = 13(2)$t$ is an exponential growth model or an exponential decay model. a. Exponential Decay Model b. Exponential Growth Model c. Both d. None of the above

#### Solution:

Substitute 0, 1, 2, and 3 for t in the exponential function y = 13(2)t.

y = 13(2)0
[Substitute 0 for t.]

= 13 x 1 = 13
[20 = 1.]

y = 13(2)1
[Substitute 1 for t.]

= 13 x 2 = 26
[21 = 2.]

y = 13(2)2
[Substitute 2 for t.]

= 13 x 4 = 52
[22 = 4.]

y = 13(2)3
[Substitute 3 for t.]

= 13 x 8 = 104
[23 = {8].]

The value of y increases by the same percent of 2 in each unit of time.

A quantity displays exponential growth if it increases by the same percent in each unit of tme.

So, the model y = 13(2)t is an exponential growth model.

Correct answer : (2)
5.
State whether the model $y$ = 8($\frac{1}{3}$)$t$ is an exponential growth model or an exponential decay model. a. Exponential Growth Model b. Exponential Decay Model c. Both d. None of the above

#### Solution:

Substitute 0, 1, 2, and 3 for t in the exponential function y = 8(1 / 3)t.

y = 8(13)0
[Substitute 0 for t]

= 8 x 1 = 8
[(1 / 3)0 = 1.]

y = 8(13)1
[Substitute 1 for t.]

= 8 x 13 = 83 = 2.66
[Simplify the fraction.]

y = 8(13)2
[Substitute 2 for t.]

= 8 x 13 x 13 = 89 = 0.88
[Simplify the fraction.]

y = 8(13)3
[Substitute 3 for t.]

= 8 x 13 x 13 x 13 = 827 = 0.29
[Simplify the fraction.]

The value of y decreases by the same percent in each unit of time.

The model y = 8(1 / 3)t is an exponential decay model.

Correct answer : (2)
6.
Victor gets a truck for $17000. The value of the truck decreases by 4% each year. Find the value of the truck after 3 years. a.$20,000 b. $14960 c.$15,738.58 d. $18,392.52 #### Solution: Let y be the price of the truck. The initial price of the truck, C is$17000.

The decay rate, r is 4% = 0.04

y = C(1 - r)t
[Write exponential decay model.]

= 17000 x (1 - 0.04)3
[Substitute 17000 for C, 0.04 for r and 3 for t.]

= 17000 x (0.96)3
[Subtract 0.04 from 1.]

= 17000 x 0.88

= 14960
[Simplify.]

The value of the truck after 3 years will be \$14960.

Correct answer : (2)
7.
State whether the model $y$ = 3($\frac{7}{3}$)$t$ is an exponential growth model or an exponential decay model. a. Exponential Growth Model b. Exponential Decay Model c. Both d. None of the above

#### Solution:

The growth or decay factor of the model y = 3(7 / 3 )t is 7 / 3 or 2.33

The growth or decay factor 2.33 > 1

The model y = 3(7 / 3)t represents an exponential growth model.

Correct answer : (1)
8.
State whether the model $y$ = 6(0.83)$t$ is an exponential growth model or an exponential decay model. a. Exponential Growth Model b. Exponential Decay Model c. Both d. None of the above

#### Solution:

The growth or decay factor of the model y = 6(0.83)t is 0.83

The growth or decay factor 0.83 < 1.

The model y = 6(0.83)t represents an exponential decay model.

Correct answer : (2)
9.
State whether the model $y$ = 8(4.09)$t$ is an exponential growth model or an exponential decay model. a. Exponential Growth Model b. Exponential Decay Model c. Both d. None of the above

#### Solution:

The growth or decay factor of the model y = 8(4.09)t is 4.09

The growth or decay factor is 4.09 >1.

The model y = 8(4.09)t represents an exponential growth model.

Correct answer : (1)
10.
What is the decay factor of the model $y$ = 6(0.7)$t$? a. 0.9 b. 0.6 c. 0.8 d. 0.7

#### Solution:

The exponential decay can be modeled by the equation y = C(1 - r)t.

The decay factor is (1 - r).

The decay factor of the model y = 6(0.7)t is 0.7
[Compare the equations.]

Correct answer : (4)

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