100 flies are put in a breeding container that can support at most 5000. If the population grows exponential..

Question

100 flies are put in a breeding container that can support at most 5000. If the population grows exponentially at a rate of 2% each day, how long will it take for the container to reach capacity?

explanations are helpful!

please explain using "e" terms : )

Answer 1

The formula for the resultant population x is
x = P(1 + r)^t

where
P = original population
r = exponential rate of growth
t = time elapsed

Therefore,
x = 5000
P = 100
r = 2% = 0.02
t = ?

Now, from the equation, divide both sides by P
x/P = (1 + r)^t

Then, get the natural logarithm of both sides
ln (x/P) = ln (1 + r)^t

Use the property of logarithms ln x^n = n ln x
ln (x/P) = t ln (1 + r)

Divide both sides by ln (1 + r). Therefore, the equation for the time t is:
t = ln (x/P)/ln (1 + r)

Substituting our values,
t = ln (5000/100)/ln (1 + 0.02)

Or
t = ln 50/ln 1.02

Or
t = 197.5507 days

^_^

Answer 2

Assume initial population a = 100 and growth p = 2% = 0.02. Then the first day you will have a + pa = a*(1+p) flies. Second day you will have a*(1+p) + p*a*(1+p) = a*(1+p)^2 flies. Third day you will have a*(1+p)^2 + p*a*(1+p)^2 = a*(1+p)^3 flies and so on. n-th day you will have a*(1+p)^n flies. Now in your case 100*1.02^n = 5000 or n*ln(1,02) = ln(50). Therefore n = 197.55 days.

Answer 3

This is basically a geometric Progression with initial term as 100 and the ratio as 1.02 (2% growth per day).

The Sum of n terms of a GP is given by,
S= a(r^n-1)/(r-1) where a is the starting term and r is the ratio. n is the number of terms (or number of days ,here.

Here a = 100 and n=1.02 and S= 5000 We have to find n

5000 = 100 (1.02^n-1/ (1.02-1)
5000*.02= 100 (1.02^n-1)
100 =100 (1.02^n-1)
1.02^n= 2
log 1.02^n=log1
n*log1.02=log1
n= log1/log1.02
=.301029996/.00860017
=35 days including the first day

Answer 4

The rate of growth is (1.02) to account for both old at 1 and new at 0.02.
Exponential growth follows this form...
Old Amount * (Growth)^ (Time) = New Amount

100 * 1.02^t = 5000
1.02^t = 50
log(1.02^t) = log(50)
t * log(1.02) = log(50)
t = log(50)/log(1.02) = 197.55 or some time in the 198th day.