Question 1199250
<pre>

The formula is

{{{A=A[o]e^(kt)}}}, where A<sub>o</sub> is the beginning amount, A is the final amount,
t is time required to reach the final amount, and k is a constant.

If we start with quantity P, then when t = 25, the final amount A is half
the starting amount P, so {{{A=expr(1/2)P}}}.

{{{expr(1/2)P=Pe^(k*25)}}}

Cancel the P's

{{{expr(1/2)cross(P)=cross(P)e^(kt)}}} 

{{{1/2=e^(k*125)}}}

Take natural logs of both sides:

{{{ln(1/2)=ln(e^(k*125))}}}

{{{-0.6931471806=k*125}}}

{{{-0.6931471806/125=k*125/125}}}

{{{-0.0055451774=k}}}

Now we go back to 

{{{A=Pe^(kt)}}}

and replace k by -0.0055451774 

{{{A=Pe^(-0.0055451774t)}}}

Now we must find how many days it will take for eighty-two percent 
of the substance to decay?

Remember, A is how much is left, not how much decaying has been
done, so when 82% of the stuff decays, what's left is 100%-82% = 18%. 
 
So we substitute 0.18P for A

{{{0.18P=Pe^(-0.0055451774t)}}}

Cancel P's

{{{0.18cross(P)=cross(P)e^(-0.0055451774t)}}}

{{{0.18=e^(-0.0055451774t)}}}

{{{ln(0.18)=ln(e^(-0.0055451774t))}}}

{{{-1.714798428=-0.0055451774t}}}

{{{(-1.714798428)/(-0.0055451774)=t}}}

{{{309.241401=t}}}

Not quite all 82% will have decayed on the 309th day,
but it will all will be decayed on the 310th day.

Edwin</pre>