Atown 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?

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.

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925

(x+3)(x+7)

step-by-step explanation: