Thread: Urgent Algebra homework help needed

1. The half-life of carbon-14 is 5700 years. If you start with 10 grams of carbon-14, how much will remain in 22,800 years?



2. A light-year is approximately 5.88 x 10 to 12 power miles. Estimate the distance from Earth to a star so far away that its light takes 9.4 years to reach us.


The answer to #1 is .625g and the answer to #2 is 5.55 x 10 to the 13th power, but I don't know how to work the problems.

Both of these problems are suppose to use exponents and I have a test on similar type questions tomorrow.

Originally Posted by chrgrl3

1. The half-life of carbon-14 is 5700 years. If you start with 10 grams of carbon-14, how much will remain in 22,800 years?

We need to use the formula:

A(t) = Ao*e^(-rt)
where A(t) is the amount remaining at time t, Ao is the initial amount, r is the rate of decrease, and t is the current time

Now, half-life = ln2/r
=> 5700 = ln2/r
=> r = ln2/5700
=> r = 0.000121604

So, using A(t) = Ao*e^(-rt), we have when t = 22800, and Ao = 10 grams:
A(t) = 10*e^(-0.000121604*22800)
......= 10*e^(-2.772588722)
......= 0.625 grams

2. A light-year is approximately 5.88 x 10 to 12 power miles. Estimate the distance from Earth to a star so far away that its light takes 9.4 years to reach us.

if it takes light 9.4 years to reach us, it means the star is 9.4 light years away.

1 light year = 5.88 x 10^12 miles
=> 9.4 light years = 9.4 x 5.88 x 10^12 miles
=> 9.4 light years = 5.53 x 10^13 miles

so the star is 5.53 x 10^13 miles from Earth

Thanks sooo much for help. Understand #2, but don't understand "In" or the logic behind formula.

Originally Posted by chrgrl3

Thanks sooo much for help. Understand #2, but don't understand "In" or the logic behind formula.

the formula i used is the standard one to use for a problem of this nature, you should pretty much memorize it, i don't remember the proof for it

ok, ln is Ln, it is called the natural log and it is defined as log to the base e. here is why the formula works out to half-life = ln2/r and r = ln2/half life

Again, using A(t) = Ao*e^(-rt)

let's assume t is the half life now. at the half life we will have half of what we started with. so let's say we started with 1 unit, then at the half life we would have 1/2 units.

so, 1/2 = 1*e^(-rt)
this is an exponential equation that we must solve using logs. since the base is e, we use the log called ln

log both sides, we get:
ln(1/2) = lne^(-rt)
=> ln1 - ln2 = lne^(-rt) ..............since log(x/y) = logx - logy
=> ln1 - ln2 = (-rt)lne ................since log(x^n) = nlogx
=> 0 - ln2 = (-rt)lne ..................since log1 = 0, no matter what base we are in
=> -ln2 = -rt ............................since lne = 1, since log[a]a = 1
=> ln2 = rt

so r = ln2/t
or t = ln2/r
where t is the half-life

got it?

Appreciate the help!

I don't know if you have had any calculus or not, but the proof of the formula used in #1 is as follows:

Let m(t) be a function that gives the mass of carbon after time 't'

Let M be the initial mass of the carbon

Let T be the half life of the carbon (i.e. the time is takes to reduce the initial mass of carbon to one half its initial mass)

Then m(0) = M and m(T) = M/2

Now the change in mass is proportional to the mass at time t; that is,
dm/dt = r * m where r is some constant to be determined.

r represents the rate at which the mass is changing.

We can no seperate the variables as follows:

dm/dt = r * m ==> dm/m = r * dt

Integrating yeilds: ln(m) = r*t + Const

Exponentiating both sides:

m = e^(r*t)*e^(Const) = C*e^(r*t) where C = e^(Const)

Now we use our initial conditions:

m(0) = M = C*e^(0) = C ==> C = M

m(T) = M/2 = M*e^(r*T) ==> 1/2 = e^(r*T) ==> r = ln(1/2)/T

Now ln(1/2) = ln(2^-1) = -ln(2) ==> r = -ln(2)/T

therefor m(t) = M * e^(-ln(2) * t/T)