We try here to calculate the general law for decay.
We have an initial population #Q_i# and, after a time interval #\Delta t# we have a new population, #Q_f#, that was the initial population times the constant of decay (that we can call #lambda#).
Mathematically we can say that
#Q_f-Q_i=-lambdaQ_i\Deltat#.
The negative sign is to take into account that the populations at the end is smaller than the population at the beginning.
We can write our equations as
#\Delta Q = -lambda Q_i\Deltat#.
The important part of this equation is that it is valid no matter how big is the time interval. If it is one hour, everything scale correctly with #-lambda Q_i# and the final result will be correct. If the time is a nanosecond it still work. Then we can consider the instantaneous change in population writing
#dQ=-lambda Q\dt#
Here I have to use #Q# instead of #Q_i# because also the initial distribution changes continuously with the time interval. Every infinitesimal time I consider I have to plug in #Q# the new initial distribution that will be changed infinitesimally from the previous.
Then I can write the differential equation
#(dQ)/dt=-lambda Q#
where the #Q# is function of time.
This is a differential equation that has the solution
#Q(t)=Q_0e^(-\lambdat)#.
Where #Q_0# is the population at the beginning of time (when #t=0#).
Now we are ready to substitute our numbers. In your case #Q_0=100#
Then the equation is #Q(t)=100e^(-lambda t)# and we also know that when #t=6# the population is the half (this is the definition of half life). So we have
#50=100e^(-lambda6)#
#1/2=e^(-lambda6)#
#ln(1/2)=-lambda6#
#lambda=-ln(1/2)/6\approx0.115#
Then the law is
#Q(t)=100e^(-0.115t)# that is the searched decay law.
