Question 741827
{{{t}}}= years since the beginning of 2000
At the beginning of 2000, {{{t=0}}} and the population is
{{{P(0)=P[0]=82000}}}
At the beginning of 2001, 1 year has passed and the population has increased by 1.6%,
meaning that the increase is {{{1.6/100=0.016}}} of the initial amount,
so it has increased by {{{0.016*P[0]=0.016*82000}}}
That means that at the beginning of 2001, when {{{t=1}}}, the population is
{{{P[0]+0.016P[0]=P[0]*(1+0.016)=P[0]*1.016}}} 
With numbers it would be {{{82000*1.016=83312}}}
That number is not important. What's important is that the population number got multiplied times {{{1.016=1+0.016}}}.
With the passage of each year the population gets increased (multiplied) by a factor of {{{1.016}}}.
After 2 years, it would have been multiplied by 1.016 again, and it would be {{{(P[0]*1.016)*1.016=P[0]*1.016^2}}}.
After 3 years, it would have been multiplied by 1.016 again, and it would be {{{(P[0]*1.016^2)*1.016=P[0]*1.016^3}}}, and so on.
If you are studying sequences and series, your teacher would say that it is a geometric sequence with a common ratio of {{{1.016}}}.
After {{{t}}} years the population will be {{{P(t)=P[0]*1.016^t}}}.
Using your initial population of {{{82000}}} at the beginning of 2000, it is
{{{highlight(P(t)=82000*1.016^t)}}}
At the beginning of 2025, {{{t=25}}} and my calculator says that
{{{P(25)=82000*1.016^25=highlight(121943)}}} (rounded)
 
NOTE: That kind of growth is called exponential growth.
If you use the growth factor per unit of time in the formula, the time will be the exponent.
In mathematics they like to use powers of the irrational number {{{e}}} instead, so you see formulas like
{{{P(t)=P[0]*e^kt}}}, but if you do not know about {{{e}}} and natural logarithms, don't worry about it, because I do not think that {{{e}}} would help your calculation at all.