Question

A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, assuming that the town grows at a constant rate of (a) 500 people per year. double once t = 10 Correct: Your answer is correct. years double twice t = 30 Correct: Your answer is correct. years double three times t = 70 Correct: Your answer is correct. years (b) 5% per year. (Round your answer to three decimal places.) double once t = 14.207 Correct: Your answer is correct. years double twice t = 28.413 Correct: Your answer is correct. years double three times t = 42.620 Correct: Your answer is correct. years

Answer 1

Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

$P_{t}=P_{0}+500t$

Where $P_{t}$ = Population after t years

$P_{0}$= Initial population

t = Time in years

a). For double once the population will be 500×2 = 10000

By plugging in the values in the equation,

10000 = 5000 + 500t

500t = 10000 - 5000

500t = 5000

t = $\frac{5000}{500}$

t = 10 years

For Double twice,

Population will be = 10000×2 = 20000

Now we plug in the values in the equation again

20000 = 5000 + 500t

500t = 20000 - 5000

500t = 15000

t = $\frac{15000}{500}$

t = 30 years

For double thrice,

Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

500t = 40000 - 5000

500t = 35000

t = $\frac{35000}{500}$

t = 70 years

b). If the population growth is 5%.

Then the growth will be exponential represented by

$T_{n}=T_{0}(1+\frac{r}{100})^{t}$

$T_{n}$ = Population after t years

$T_{0}$ = Initial population

t = time in years

For double once,

Population after t years = 10000

$10000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{10000}{5000}$

$(1.05)^{t}=2$

Take log on both the sides

$log(1.05)^{t}=log2$

tlog(1.05) = log2

t = $\frac{log2}{log1.05}$

t = 14.20 years

For double twice,

Population after t years = 20000

$20000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{20000}{5000}$

$(1.05)^{t}=4$

Take log on both the sides

$log(1.05)^{t}=log4$

tlog(1.05) = log4

t = $\frac{log4}{log1.05}$

t = 28.413 years

For double thrice

Population after t years = 40000

$40000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{40000}{5000}$

$(1.05)^{t}=8$

Take log on both the sides

$log(1.05)^{t}=log8$

tlog(1.05) = log8

t = $\frac{log8}{log1.05}$

t = 42.620 years