Question 194462
A radioactive substance has a half-life of 200 years. We currently have 15 grams of the substance. 

How long until only .5 grams of the substance remains radioactive? 

Could you please provide a detailed explanation of how you achieved the solution. Thanks!

The exponential growth and decay formula is

{{{A = Pe^(r*t)}}}

Where A = the final amount
      P = the beginning amount
      r = the rate (positive for growth, negative 
          for decay. This is a decay problem so
          we expect r to be negative)
      t = the time that has lapsed 
      e = 2.718...

When t = 0, A = 15

{{{15 = Pe^(r*0)}}}

{{{15 = Pe^0}}}

{{{15=P(1)}}}

{{{15=P}}}

Substitute 15 for P in 

{{{A = 15e^(rt)}}}

Since the substance has a half-life of 200
years, then there will only be half of 15
grams present, so

when t = 200, A = 7.5, half of 15 grams.

So we substitute that and get

{{{7.5 = 15e^(r*200)}}}

{{{7.5/15 = e^(r*200)}}}

{{{.5 = e^(200r)}}}

Use the fact that the equation {{{Y=e^X}}} 
is equivalent to {{{X =ln(Y)}}}

{{{200r=ln(.5)}}}

{{{r = ln(.5)/200}}}

{{{r = -.0034657359}}}

Will finish later.