Question 998642
General model  {{{y=pe^(-kt)}}} and you have starter information so that you can find the value for k.  Also half-life.


The 52% decrease after 980 days,
{{{y/p=e^(-kt)}}}
{{{ln(y/p)=-kt}}}
{{{ln(p/y)=kt}}}-----------------------(and this step you will use again).
{{{highlight(k=(1/t)ln(p/y))}}}
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Information given in description is {{{system(t=980*days,p=100,y=48)}}}, and this "48" is because that is what remains as percent after 52% decayed.


You should be able to do the rest of the questions and steps.


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Still wanted more steps to the solutions---


(a) Half-Life

Return to this part: {{{ln(p/y)=kt}}}
What was k?  Substitute the given values in the main part of the description!
{{{k=ln(100/48)/980}}}
{{{highlight(k=7.489*10^(-4))}}}
The model for decay is now {{{highlight_green(y=pe^(-0.0007489*t))}}}.
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The step to find any time t,
{{{t=ln(p/y)/k}}}
You know the needed values for finding half-life with this formula, so substitute THEM.
{{{t=ln(1/(1/2))/0.0007489}}}
{{{highlight(t=925*days)}}}----the half-life.


(b)  How long to decay from 100 mg to 77 mg ?
You have the formula and model to use.