Question 837970
{{{y}}}= amount of dye left in Steven, in mg, at time {{{t}}} in hours.
{{{A}}}= original amount of the dye, in mg, that he took, and had inside at {{{t=0}}} .
 
THE (precocious) FIFTH GRADER THOUGHTS:
Since the half-life of the dye is 2 hours,
after {{{red(1)}}} half-life, at {{{t/2=red(1)}}} , and
{{{y=A*(1/2)}}} is half of the original amount
After {{{red(2)}}} half-lives, {{{t/2=red(2)}}} another half-life has passed, and
{{{y=(A*(1/2))*(1/2)=A*(1/2)^red(2)}}} .
Another half-life after that, after {{{red(3)}}} half-lives, {{{t/2=red(3)}}} , and
{{{y=(A*(1/2)^2)*(1/2)=A*(1/2)^red(3)}}} .
The pattern tells me that at {{{red(t/2)}}}{{{half-lives}}} the amount left is
{{{y=A*(1/2)^red(t/2)=A/2^red("t / 2")}}} .
So if {{{20=A/2^"t / 2"}}} , {{{A=20*2^"t / 2"}}} .
Since {{{t=17}}} , {{{t/2=17/2=8&1/2}}} , and {{{2^"t / 2"=2^(8+"1 / 2")=2^8*sqrt(2)=256sqrt(2)="approx ."}}}{{{256*1.414}}} .
A good approximate solution is
{{{A=20*256*1.414=7239.68="approx ."}}}{{{highlight(7240)}}} .
Steven took in 7240 mg (7.24 grams) of red dye.
 
THE PRE-CALCULUS STUDENT THOUGHTS:
I know the teacher said this type of problem is an example of exponential decay, and there is a formula for this.
It is an exponential function, with that irrational letter {{{e}}} and a complicated exponent.
There was {{{t}}} in that exponent, and something else, maybe the half-life, or some strange constant, or both.
 
THE PRE-CALCULUS STUDENT WITH HELP FROM THE FIFTH GRADER:
I do mot enjoy memorizing and blindly applying formulas.
I keep forgetting those formulas, anyway.
{{{y=A/2^red("t / 2")}}}
That {{{red("t / 2")}}} is the number of half-lives,
because in this problem the half-life is {{{t["1/2"]=red(2)}}} ,
and so {{{t/t["1/2"]=red(t/2)}}} is the number of half-lives.
{{{ln(y)=ln(A/2^red("t / 2"))}}}
{{{ln(y)=ln(A)-ln(2^red("t / 2"))}}} --> {{{ln(y)/ln(A)=-red(t/2)*ln(2)}}} --> {{{ln(y)-ln(A)=-ln(2)*red(t/2)}}} --> {{{ln(y/A)=-ln(2)*(t/t["1/2"])}}} --> {{{ln(y/A)=-ln(2)*t/t["1/2"]}}} --> {{{ln(y/A)=-(ln(2)/t["1/2"])*t}}}
I think that factor was called {{{lambda}}} or {{{k}}} ,
{{{lambda=ln(2)/t["1/2"]}}} ,
and you can write
{{{ln(y/A)=-lambda*t}}} --> {{{y/A=e^(-lambda*t)}}} --> {{{y=A*e^(-lambda*t)}}}
Oh, and we were supposed to use the approximate value {{{ln(2)=0.693}}} ,
so {{{lambda=0.693/2-0.3465}}} for this problem, so {{{y=A*e^(-0.3465*t)}}}.
At {{{t=17}}} , we have {{{20=A*e^(-0.3465*17)}}}
Solving for {{{A}}} ,
{{{A=20/e^(-0.3465*17)=20/e^(-5.8905)=20/0.00276559}}}(rounding)
So {{{A=7232}}} (rounded).
What about using a better approximation for {{{ln(2)}}} ?
With all the digits my calculator carries, {{{A=7241}}} (rounded).