Question

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?

Answer 1

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%`).