Question 1058294
b will contain the decay factor if you are using that formula shown.  You have two points,  (t,P).  Try re-handling the variable so you have t=0 for year 2006 and t=2 for year 2008.  Your two points are  (0,15574) and (2,11788).


You choose the base you want, but  {{{log((P))=log((P[o]))+log((b^t))}}}


{{{log((P))=t*log((b))+log((P[o]))}}}, but not sure if this is necessary.


Look at the second point:
{{{11788=15574*b^2}}}
{{{b^2=11788/15574}}}
{{{b^2=0.7569026}}}
{{{highlight(b=0.87)}}} for model as {{{highlight(P(t)=15574(0.87)^t)}}}.


You can break the b into its percents pieces as {{{1.00-0.13}}}.  This is a yearly decrease of 13%.


The four-year value would be the evaluation  {{{P(4)=15574*(0.87)^4}}}.