Is the equation #A=21000(1-.12)^t# a model of exponential growth or exponential decay, and what is the rate (percent) of change per time period?

1 Answer
May 9, 2016

This equation describes a decay since
#0 < (1-.12)=0.88 < 1#.
At #t=0# its value is #A=21000#. As #t->oo#, the value asymptotically diminishes to #0#.
Percent of change is #12%# per unit of time.

Explanation:

Consider a function #f(x)=a*q^x# with #a > 0# and #0 < q < 1#
For #a=2# and #q=0.9# the graph of this function is below:

graph{2(.9)^x [-10, 10, 5, -5]}

At #x=0# the function value is #f(0)=a>0#
As #x->oo#, since #0 < q < 1#, we multiply #a# by progressively smaller and smaller number #q^x->0#.
The result, therefore, will be asymptotic behavior #f(x)->0#

Therefore, ant function #f(x)=a*q^x# with #a > 0# and #0 < q < 1# describes decay.
The value of this function is diminishes during the time interval from #t=N# to #t=N+1# by a factor of #q#, which is the same as to state that its value diminishes from #a*q^N# to #a*q^(N+1)#. The difference between old and new values is
#a*q^N - a*q^(N+1) = a*q^N(1-q)#
So, #1-q# constitutes the rate of change, which in many cases is expressed as percentage.

For #a=21000# and #q=1-.12# this rate of change is
#1-q = 1 - (1-.12) = .12# (or #12%#).