a population doubles every month
a-initial population 20, what is function that represents poulation with respect to time
b-how many present after 1 year and 5 years
c-when will there be 10,000
a) number = 20 (2)^(t) , where t is in months
b) in 1 year , t = 12
number = 20(2^12) = 81920
in 5 years , t = 60
number = 20(2^60) = 2.306 x 10^19 , ridiculous answer !
c) 20(2^t) = 10000
2^t = 500
ln both sides
tln2 = ln500
t = ln500/ln2 = 8.965 monts or appr 9 months
after m months,
p = 20*2^m
plug in values for m and evaluate
solve 20*2^m = 10000: m=8.
a) In order to determine the function that represents the population with respect to time, we can use the formula for exponential growth. The general form of this formula is:
P(t) = P0 * (1 + r)^t
Where:
- P(t) represents the population at time t
- P0 represents the initial population
- r represents the growth rate
- t represents the time elapsed
Since the population doubles every month, the growth rate r can be calculated as the natural logarithm of 2 (ln(2)) divided by the number of months in a year (12), to convert it from monthly growth to annual growth. So we have:
r = ln(2) / 12
Substituting the given values into the formula, we get:
P(t) = 20 * (1 + ln(2) / 12)^t
b) To find the population after 1 year and 5 years, we substitute the corresponding values of t into the formula.
P(1) = 20 * (1 + ln(2) / 12)^1
P(5) = 20 * (1 + ln(2) / 12)^5
c) To determine when the population will reach 10,000, we can rearrange the formula and solve for t:
10,000 = 20 * (1 + ln(2) / 12)^t
To solve for t, we can take the natural logarithm of both sides to eliminate the exponent:
ln(10,000) = ln(20 * (1 + ln(2) / 12)^t)
After simplifying, we have:
ln(10,000) = ln(20) + ln(1 + ln(2) / 12)^t
Now, we can isolate t by dividing both sides by ln(1 + ln(2) / 12):
t = ln(10,000) - ln(20) / ln(1 + ln(2) / 12)
Using a calculator or a math software, you can evaluate the expression on the right side to find the value of t.