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. (b) 5% per year. The growth of an animal population, P, is described by the function P=300·2t/20. (a) How large is this population in year t=0? t=20? (b) When does this population reach 1000?

Answer: Hello there!

first question!

at t = 0 we have a population of 5000.

we want to see how long it takes for the population to double once, twice and three times if:

a) the rate of growth if by 500 people by year:

then the equation of growth is

p (t) = 5000 + 500t where t is years.

now we are seeking for

p (x) = 5000 + 500x = 2*5000 = 10000

x = (10000 - 5000) / 500 = 10

you need 10 years in order to double the population one time.

for the second time:

p (y) = 5000 + 500y = 2*10000 = 20000

y = (15000) / 500 = 30 more years to double twice

the third time:

p (z) = 5000 + 500z = 2*20000 = 40000

z = 35000/500 = 70 years to double three times

b) 5% per year

here the equation is P (t) = 5000 * (1.05) ^ (t) where again t is years. and the 1.05 apears because if the year zero the population is 100%, the next year there are a 105%, and so on.

for double it once we have

p (x) = 5000 * (1.05) ^ (x) = 10000

(1.05) ^ (x) = 10000/5000 = 2

now we need to find the value of t.

ln ((1.05) ^ (x)) = ln (2)

x*ln (1.05) = ln (2)

x = ln (2) / ln (1.05) = 14.2 years

This rate starts slower than the first

for double twice we need to see:

p (y) = 5000 * (1.05) ^ (y) = 2*10000 = 20000

y = ln (4) / ln (1.05) = 28.4

and now is faster than the first.

for it to double three times:

p (z) = 5000 * (1.05) ^ (z) = 2*20000 = 40000

z = ln (8) / ln (1.05) = 42.6 years

second problem:

we have P=300·2t/20, which is weirdly written, but ok, let's solve it.

when t = 0, we replace the value of t in the equation, and we get

p = 300·2*0/20 = 0

when t = 20

p = 300·2*20/20 = 300.2

when does this population reach 1000?

p (x) = 300.2x/20 = 1000

300.2x = 20*1000 = 20000

x = 20000/300.2 = 66.6 years.

Now we can try with another way if writting this that has more sense:

P (t) = 300 * (2^ (t/20))

again, when t = 0

P (0) = 300 * (2^ (0/20)) = 300

when t = 20

P (20) = 300 * (2^ (20/20)) = 300*2 = 600

and we want to finde the number of years x needed to have a population equal to 1000:

p (x) = 300 (2^ (x/20)) = 1000

(2^ (x/20)) = 1000/300 = 3.33

aply ln to both sides:

Ln (2) * x/20 = ln (3.33)

x = ln (3.33) * 20/ln (2) = 34.7 years.