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.
