law of growth and decay problem?

Question

1. the population of a city doubles in 50 years, how many years will it be four times as much? Assume that the rate change is proportional to the numbers of inhabitants.
ans. 120 years

2. a certain substance decomposes at a rate proportional to the amount present. If one-third of it disapears after 1000 days, what is the percentage lost in 100 days?
ans. 3.97 %

3. thirty percent of a radioactive substance disappears in 15 years.Find the half-life of the substance.
ans 29.12 years

4. sugar decomposes in water at a rate proportional to the amount still unchanged.If there were 50 lb of sugar present initially amd at the end of 5 hours this is reduced to 20 lb, how long will it take until 90% of the sugar is decomposed?

Please help me solve this problem...i can't find the solution
please.........

thanks a lot

Answer 1

1)
rate of change is proportional to number of inhabitants
Let say x = number of inhabitants

dx/dt = kx
1/x dx = k dt
ln x = kt + C where C is constant

Let say orginal x = P
ln P = C

population doubles in 50 years
ln 2P = k(50) + ln P
50 k = ln 2
k = 1/50 ln 2

ln x = [1/50 ln 2]t + ln P
For population become 4 times as much as original
ln 4P = [1/50 ln 2]t + ln P
[1/50 ln 2]t = ln 4
t = 100 years

2)
Rate of decomposition is propotional to amount present
dx/dt = kx
1/x dx = k dt
ln x = kt + C where C is constant

Let say original amount = A
ln A = C

1/3 substance disappear after 1000 years, left 2/3 substance
ln 2/3A = k(1000) + ln A
k = 1/1000 ln 2/3

After 100 days
ln x = [1/1000 ln 2/3] t + ln A
ln x = [1/1000 ln 2/3](100) + ln A
ln x = ln (2/3)^(1/10) A
x = 0.96 A

Percentage lost
= (A - 0.96A / A) x 100%
= 3.97 %
3)
30% of radioactive substance disappear in 15 years,
so only left 70% substance
Let say original = A
0.7A = A (1/2^(15/n))
0.7 = (1/2^(15/n))
ln 0.7 = (15/n) ln (1/2)
15/n = ln 0.7 / ln (1/2)
15/n = 0.5145
n = 29.15 years half life

4)
Sugar decomposes at rate proportional to amount still unchanged
Initial sugar amount = 50 lb
After 5 hours, reduced to 20 lb
Let say x = amount of sugar
dx/dt = kx
1/x dx = k dt
ln x = kt + C where C is constant

When t=0, x = 50 lb
ln 50 = C

When t=5 hrs, x = 20 lb
ln 20 = k(5) + ln 50
5k = ln (20/50)
k = 1/5 ln (2/5)

ln x = [1/5 ln (2/5)] t + ln 50

If 90% sugar decomposed, left 10% sugar.
ln (1/10 x 50 lb) = [1/5 ln (2/5)] t + ln 50
ln 5 - ln 50 = [1/5 ln (2/5)] t
ln (1/10) = [1/5 ln (2/5)] t
t = 12.56 hours