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.