# Thread: Exponential Growth and Decay

1. ## Exponential Growth and Decay

I'm having some trouble with the following problems:

1) Radioactive decay problem: What percent of a present amount of radioactive radium (226Ra) will remain after 900 years? (Half-life: 1599 years)

2) Radioactive decay problem: Find the half-life of a radioactive material if after 1 year 99.57% of the initial amount remains.

3) In 1990, the population of Mexico was 89 million and growing at the rate of 1.8% per year. In 1990, the US population was 250 million and growing at the rate of .7% per year. If these rates continue, when will the 2 nations have the same population?

2. Originally Posted by cupcakelova87
I'm having some trouble with the following problems:
For radioactive decay we use the formula:

A(t) = A0*e^(-rt)

where A(t) is the amount left after time t, A0 is the intitial amount, r is the rate of decrease, t is the time

1) Radioactive decay problem: What percent of a present amount of radioactive radium (226Ra) will remain after 900 years? (Half-life: 1599 years)
r = ln2/t, where t is the half-life
=> r = ln2/1599 = 0.000433487

we want A(t) when t is 900, that is, we want A(900)

A(900) = A0 *e^(-0.000433487(900)) = A0(0.676963)

so A(900) = (0.676963)A0

so A(900) is 67.69% of A0. this is the percent remaining

2) Radioactive decay problem: Find the half-life of a radioactive material if after 1 year 99.57% of the initial amount remains.
assume we start with 1 unit of the radioactive substance, then after 1 year, 0.9957 units are left.

using A(t) = A0*e^(-rt)

we have A(1) = 0.9957 = 1*e(-r(1))
=> e^-r = 0.9957
=> ln(e^-r) = ln(0.9957)
=> -r = ln(0.9957)
=> r = - ln(0.9957)
=> r = 0.004309271

now t = ln2/r, where t is the half life
=> t = ln2/0.004309271
=> t = 160.8502 years

3. Originally Posted by cupcakelova87

3) In 1990, the population of Mexico was 89 million and growing at the rate of 1.8% per year. In 1990, the US population was 250 million and growing at the rate of .7% per year. If these rates continue, when will the 2 nations have the same population?
For exponential growth, we use the same formula for decay, but with one difference, the power of e is positive. that is, we use the formula:

P(t) = P0*e^(rt)

where P(t) is the population at time t, P0 is the initial population, r is the rate of growth, t is the time

now for mexico, we are given r = 0.018 and the initial population in 1990 is P0 = 89 million.
that is, for Mexico, P(t) = 89000000e^0.018t

now for the US, we are given r = 0.007 and the initial population in 1990 is 250 million
that is, for the US, P(t) = 250000000e^0.007t

when will they have the same population?
when 89000000e^0.018t = 250000000e^0.007t of course
now all we do is solve for t
=> e^0.018t = (250/89)e^0.007t ..........i divided both sides by 89 million
=> e^0.018t/e^0.007t = 250/89 ...........i divided both sides by e^0.007t
=> e^0.011t = 250/89 ........................when dividing the same base we subtract the powers. 0.018t - 0.007t = 0.011t

now we have an exponential equation. how do we solve this? we log both sides of course

=> lne^0.011t = ln(250/89)
=> 0.011t = ln(250/89)
=> t = ln(250/89)/0.011
=> t = 93.89 ~= 94 years

so the populations will be equal after about 94 years, that is in the year 2084

4. Thanks Jhevon! I have one more problem...

On the Richter scale, the magnitude R of an earthquake of intensity I is given by R=(ln I - ln Io)/ln 10
where Io is the minimum intensity used for comparison. Assume Io=1.
a) Find the intensity of the 1906 SF earthquake in which R=8.3.
b) Find the factor by which the intensity is increased when the value of R is doubled.
c) Find dR/dI.

5. Sorry, I never noticed this question 'till just now

Originally Posted by cupcakelova87
Thanks Jhevon! I have one more problem...

On the Richter scale, the magnitude R of an earthquake of intensity I is given by R=(ln I - ln Io)/ln 10
where Io is the minimum intensity used for comparison. Assume Io=1.
a) Find the intensity of the 1906 SF earthquake in which R=8.3.
If Io = 1, then ln(Io) = 0
So we have R = ln(I)/ln(10)

=> ln(I) = ln(10)R
=> I = e^(R*ln(10)) = e^ln(10^R) = 10^R
When R = 8.3
I = 10^8.3 ~= 1.995 x 10^8

b) Find the factor by which the intensity is increased when the value of R is doubled.
when R = 2*8.3 = 16.6
I = 10^16.6 ~= 3.981 x 10^16

so I increased by a factor of (3.981 x 10^16)/(1.995 x 10^8) ~= 1.995 x 10^8 ~= 2 x 10^8

so basically, I is squared when R is doubled

c) Find dR/dI.
R = ln(I)/ln(10)
=> dR/dI = (1/ln(10))*(1/I) = 1/ln(10)*I