SOLUTION: If 80% of a radioactive element remains radioactive after 200million years, then what percentage remains radioactive after 600million years? and also, what is the half life of thi

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Question 1070297: If 80% of a radioactive element remains radioactive after 200million years, then what percentage remains radioactive after 600million years? and also, what is the half life of this element?
I have the correct answers: 51% and 621 million years
(this is a quiz question that i failed but need to understand how to do it for the next attempt)

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
A=Pe%5E%28RT%29

where 
      P = the beginning amount
      T = the number of years
      R = the rate of decay
      A = amount left after T years.

In the first case we are given that
when T = 200000000=2*10^8, A = (80% of P) = 0.8P

So we substitute 200000000 for T and 0.8P for A

0.8P=Pe%5E%28R%28200000000%29%29

P's cancel on both sides:

0.8=e%5E%28R%28200000000%29%29

Take natural logarithms of both sides:

ln%280.8%29=ln%28e%5E%28R%28200000000%29%29%29

ln%280.8%29=+R%28200000000%29

ln%280.8%29%2F%28200000000%29=R

Do that on a calculator and get 

-1.11571776%2A10%5E%28-9%29=+R

So now the formula 

A=Pe%5E%28RT%29

becomes

matrix%282%2C1%2C%22%22%2CA=Pe%5E%28%28-1.11571776%2A10%5E%28-9%29%29T%29%29

And since we want to know what percentage of the
original amount P, we let x be the decimal equivalent 
that we must multiply the original amount P by, so
we substitute xP for A and T=600000000 = 6x108

matrix%282%2C1%2C%22%22%2CxP=Pe%5E%28%28-1.11571776%2A10%5E%28-9%29%29600000000%29%29

Cancel the P's and add simplify the exponent

matrix%282%2C1%2C%22%22%2Cx=e%5E%28%28-6.69430656%2A10%5E%28-1%29%29%29%29
  
matrix%282%2C1%2C%22%22%2Cx=e%5E%28-0.669430656%29%29

matrix%282%2C1%2C%22%22%2Cx=0.5119999989%29

That rounds to 0.51 or 51%

--------------------------------------

The half-life is when A = one-half of the original amount P

So we substitute expr%281%2F2%29P for A

matrix%282%2C1%2C%22%22%2CA=Pe%5E%28%28-1.11571776%2A10%5E%28-9%29%29T%29%29

matrix%282%2C1%2C%22%22%2Cexpr%281%2F2%29P=Pe%5E%28%28-1.11571776%2A10%5E%28-9%29%29T%29%29

The P's cancel

matrix%282%2C1%2C%22%22%2Cexpr%281%2F2%29=e%5E%28%28-1.11571776%2A10%5E%28-9%29%29T%29%29

Take natural logs of both sides:

  

matrix%282%2C1%2C%22%22%2Cln%281%2F2%29=%28-1.11571776%2A10%5E%28-9%29%29T%29

matrix%282%2C1%2C%22%22%2Cln%281%2F2%29%2F%28-1.11571776%2A10%5E%28-9%29%29=T%29

Do that on a calculator and get

matrix%282%2C1%2C%22%22%2C621256742=T%29

which rounds to 621000000 or 621 million years.

[Note: this would be easier if you would use the STO
key on your TI calculator to store the messy values,
so you could just use the letters. Maybe you just
round off as you go.]

Edwin