Question 772377
What's probably expected is blind application of a memorized formula involving {{{e}}} while using a calculator and talking about exponential functions and exponential decay.
 
THE FANCY WAY:
Exponential decay functions state that the amount {{{A}}} remaining after a time {{{t}}} is related to the initial amount {{{A[0]}}} and the half-life {{{t["1/2"]}}}} by the function
{{{ln(A)=ln(A[0])-(ln(2)/t["1/2"])*t}}} or the corresponding exponential {{{A=A[0]*e^(-(t*ln(2)/t["1/2"]))}}}
Substituting the values given,
{{{ln(5)=ln(x)-(ln(2)/25)*100}}}-->{{{ln(5)=ln(x)-ln(2)*(100/25)}}}-->{{{ln(5)=ln(x)-4*ln(2)}}}-->{{{ln(x)=ln(5)+4*ln(2)}}}
{{{ln(x)=ln(5)+4*ln(2)}}-->{{{ln(x)=ln(5)+ln(2^4)}}}-->{{{ln(x)=ln(5)+ln(16)}}}-->{{{ln(x)=ln(16*5)}}}-->{{{ln(x)=ln(16*5)}}}--> {{{highlight(x=80)}}}
 
THE CONCEPTUAL, MENTAL MATH WAY:
However, {{{100}}} days is {{{100/25=4}}} half lives, and that means the amount {{{x}}} was halves 4 times, and is now {{{1/16}}} of the original amount.
Doubling the final 5g amount 4 times, gives us the original amount as
{{{highlight(x=80g)}}}
The calculation can be done without touching pencil or calculator.
5 --> 10 --> 20 --> 40 --> 80