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.