Given:-
[tex]v(t)=21300(1.24)^t[/tex]
To find the initial value, does the fucnction represent growth or decay.
Here the value of a is 21300.
So we get,
[tex]1+r=1.24[/tex]
So the value of r is,
[tex]r=0.24[/tex]
So now we get percentage is,
[tex]24\%[/tex]
Rate of change is 24%.
So the required solution is,
[tex]initial\text{ value =21300}[/tex]
So the percentage range is,
[tex]24\%[/tex]
