Question 1202012: If 80% of a radioactive element remains radioactive after 200 million years, then what percent remains radioactive after 700 million years? What is the half life of this element?
Found 4 solutions by ikleyn, Theo, greenestamps, josgarithmetic: Answer by ikleyn(52814) (Show Source): Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! formua you can use is f = p * (1+r) ^ n
f is the future value
p is the present value
r is the growth rate per time period.
2 + r is the growh factor per time period.
n is the number of time periods
the time periods are in millions of years.
after 200 million years, 80% of the element remains radioactive.
formula becomes:
.8 = 1 * (1 + r) ^ 200
divide both sides of the equation by 1 to get:
.8/1 = (1 + r) ^ 200
solve for (1 + r) to get:
(1 + r) = (.8/1) ^ (1 / 200) = .9988849044
that says that the growth factor is .9988849044 every million years.
to confirm, replace n in the original equation and solve for f to get:
f = 1 * .9988849044 ^ 200 = .8
now that you have the growth factor for every 1 million years, you can solve for the remaining percent after 700 million years.
the formula becomes f = 1 * .9988849044 ^ 700 = .4579467218 = 45/80%.
to find the half life, set f = .5 in the original equation and solve for n.
you will get:
.5 = 1 * .9988849044 ^ n
divide both sides of the eqution by 1 to get:
.5/1 = .9988849044 ^ n
simplify to get:
.5 = .9988849044 ^ n
take the log of both sides of the equation to get:
log(.5) = log(.9988849044 ^ n)
by log rule that says log(x^n) = n * log(x), this becomes:
log(.5) = n * log(.9988849044)
divide both sides of this equation by log(.9988849044) to get:
log(.5) / log(.9988849044) = n
solve for n to get:
n = 621.2567439 million years.
to confirm, replace n in the original eqution and solve for f to get:
f = 1 * .9988849044 ^ 621.2567439 = .5
the equation can be graphed as shown below:

Answer by greenestamps(13203) (Show Source): Answer by josgarithmetic(39620) (Show Source):
|
|
|