Exponential Models Worksheet Answer Key – Algebra 2

Problem 1 Growth or decay

a) Factor 1.04 > 1: growth of 4%.

b) Factor 0.85 < 1: decay of 15%.

c) Factor 2: growth of 100%.

d) Factor 0.97: decay of 3%.

Problem 2 A growing town

The growth factor is 1 + 0.03 = 1.03, so P(t) = 8,400(1.03)^t.

P(5) = 8,400(1.03)⁵ ≈ 9,738.

P(t) = 8,400(1.03)^t; about 9,738 people

Problem 3 A depreciating car

The decay factor is 1 − 0.11 = 0.89, so V(t) = 31,500(0.89)^t.

V(3) = 31,500(0.89)³ ≈ $22,206.52.

V(t) = 31,500(0.89)^t; about $22,206.52

Problem 4 Reading an exponential graph

Initial value a = 200 and growth factor b = 1.25, which is 25% growth per period.

y(2) = 200(1.5625) = 312.50. y(3) = 200(1.953125) = 390.63.

Initial value 200; factor 1.25; y(2) = 312.50; y(3) = 390.63

Problem 5 Evaluating powers of e

a) e^0.5 ≈ 1.649.

b) e⁻¹ ≈ 0.368.

c) e^2.3 ≈ 9.974.

1.649; 0.368; 9.974

Problem 6 Annual compounding

A = 2,500(1 + 0.045)⁶ = 2,500(1.045)⁶ ≈ $3,255.65.

$3,255.65

Problem 7 Monthly compounding

n = 12 and nt = 60: A = 4,000(1 + 0.036/12)⁶⁰ = 4,000(1.003)⁶⁰ ≈ $4,787.58.

$4,787.58

Problem 8 Continuous compounding

A = 1,500e^{0.05 · 8} = 1,500e^0.4 ≈ $2,237.74.

$2,237.74

Problem 9 Comparing compounding frequencies

Annually: 1,000(1.06)¹⁰ ≈ $1,790.85.

Quarterly: 1,000(1.015)⁴⁰ ≈ $1,814.02.

Monthly: 1,000(1.005)¹²⁰ ≈ $1,819.40.

Continuously: 1,000e^0.6 ≈ $1,822.12.

More frequent compounding gives more, but the differences are small.

$1,790.85; $1,814.02; $1,819.40; $1,822.12

Problem 10 Interpreting parameters

15,000 is the initial value of the van, in dollars. The factor 0.9 means the value is multiplied by 0.9 each year, a loss of 10% per year.

V(2) = 15,000(0.81) = $12,150.00.

Initial value $15,000; loses 10% yearly; $12,150.00

Problem 11 Half-life

After 18 days there are 3 half-lives: 80 ÷ 2 ÷ 2 ÷ 2 = 10. Check: 80(1/2)³ = 10.

After 3 days: 80(1/2)^0.5 = 80/√2 ≈ 56.57.

10 mg after 18 days; about 56.57 mg after 3 days

Problem 12 Doubling time

N(t) = 500 · 2^t/3.

After 12 hours: 500 · 2⁴ = 8,000.

After 5 hours: 500 · 2^5/3 ≈ 1,587.

N(t) = 500 · 2^t/3; 8,000 cells; about 1,587 cells

Problem 13 Changing the time unit

(1.5)^t/2 = (1.5^1/2)^t, so b = √1.5 ≈ 1.2247.

The model is y = 200(1.2247)^t, an increase of 22.5% per year.

y ≈ 200(1.2247)^t; about 22.5% per year

Problem 14 Model from a table

Ratios: 460/400 = 1.15, 529/460 = 1.15, 608.35/529 = 1.15. A constant ratio means exponential growth.

15% growth per year; V(t) = 400(1.15)^t

Problem 15 Doubling an investment

A(t) = 5,000(1.06)^t.

t = 10: $8,954.24.

t = 11: $9,491.49.

t = 12: $10,060.98.

The balance first reaches $10,000 during year 12.

12 years

Problem 16 A shrinking town

P(t) = 12,500(0.975)^t.

t = 7: 10,470.

t = 8: 10,208.

t = 9: 9,953.

P(t) = 12,500(0.975)^t; below 10,000 in year 9

Problem 17 Carbon dating

11,460 = 2 × 5,730, so there are 2 half-lives: (1/2)² = 1/4.

2,865 is half of a half-life: (1/2)^0.5 ≈ 0.7071, about 70.7%.

One quarter; about 70.7%

Problem 18 Effective annual yield

Monthly: (1.005)¹² − 1 ≈ 0.06168, so the yield is about 6.168%.

Continuous: e^0.06 − 1 ≈ 0.06184, about 6.184%.

Monthly 6.168%; continuous 6.184% (slightly higher)

Problem 19 Choosing between two offers

A: 3,000(1 + 0.05/4)²⁸ = 3,000(1.0125)²⁸ ≈ $4,247.98.

B: 2,950e^{0.052 · 7} = 2,950e^0.364 ≈ $4,245.27.

Offer A is greater: $4,247.98 compared with $4,245.27

Problem 20 Model from two data points

Divide the two equations: ab³ab = b² = 399.3/330 = 1.21, so b = 1.1.

Then a(1.1) = 330 gives a = 300.

y(5) = 300(1.1)⁵ = 483.153.

a = 300, b = 1.1; y(5) ≈ 483.15