Question 845641
A After one half-life the fraction of the element that remains is {{{1/2}}} .
After {{{2}}} half-lives the fraction of the element that remains is {{{1/2}}} of {{{1/2}}} , which is {{{(1/2)*(1/2)=(1/2)^2}}} .
After {{{3}}} half-lives the fraction of the element that remains is {{{(1/2)^3}}} .
The same idea works for any number of half-lives, even if that number is not an integer.
{{{t}}} years is {{{t/6500}}} half-lives,
After that time, the fraction of the element that remains is
{{{(1/2)^"t / 6500"}}}
As a percentage, it is {{{f(t)=100*(1/2)^"t / 6500"}}} .
That is an exponential function with base {{{1/2}}} , and that is not a fashionable base.
Calculators have exponential functions with base {{{10}}} and with base {{{e}}} .
The most popular base for exponential functions is the irrational number {{{e}}} .
The reason for that is that calculus works better with {{{e}}} as a base,
so even if you do not intend to ever study calculus, they make you use {{{e}}} rather than {{{10}}} .
It is customary to write the function as
{{{f(t)=100*e^("- ln ( 2 ) * t / 6500")}}} or  {{{f(t)=100*e^("- 0.693*t / 6500")}}} .
It is really the same thing, because the natural logarithm of {{{(1/2)^"t / 6500"}}} is
{{{(t/6500)*ln(1/2)=(t/6500)*(-ln(2))=-ln(2)*t/6500}}} .
People memorize the "formula" for the exponential function with base {{{e}}} without understanding it.
I never memorized that formula, but I can deduce the one with {{{1/2}}} as a base from the definition of half-life, and then I can "translate" it to base {{{e}}} .
 
B When {{{t=100}}} , {{{-0.693*t/6500=-69.3/6500}}} and
{{{f(t)=100*e^("-69.3 / 6500")=100*0.989=98.9}}}
So after 100 years 98.9% of the radioactive element remains.
 
C {{{26000/6500=4}}} so 26,000 years is {{{4}}} half-lives.
The fraction that remains after {{{4}}} half-lives is
{{{(1/2)^4=1/16=0.0625}}} That is 6.25% .
No formula used. We do not need {{{e}}} for that.