Question 1090926
{{{y=12053*1.0345^x}}} where
{{{y}}}= population of the town, and
{{{x}}}= number of years after the year 2000. 
 
1. In the year 2000, {{{x=0}}}
{{{1.0345^0=1}}} , because a non-zero number to the zero power is by definition equal to {{{1}}} .
So, in the year 2000,
{{{y=12053*1.0345^0=y=12053*1=highlight(12053)}}} .
 
2. {{{highlight(y=12053*e^(ln(1.0345)*x))}}} is probably the expected/accepted answer,
although {{{highlight(y=12053*e^(0.033918*x))}}} is a good approximation,
and {{{highlight(y=12053*1.0345^x)}}} is a valid answer.
 
NOTE:  
The teacher may not accept any true answer not fitting what was taught in class. That may include formulas, symbols, and formats that I can an only guess at. If you are asked to "show your work", you may be expected to start by writing a general "formula" for an exponential growth model, such as
{{{P=P[0]*e^rt}}} or {{{y=A*e^kx}}} (Letter symbols may vary).
FURTHER EXPLANATION:
In this problem, the word "model" refers to a function/equation (such as as {{{y=12053*1.0345^x}}} ) that fits fairly well the data available, and may be a useful "model" to predict population growth in the near future, or estimate what the population was at a time there is no data for.
The word "continuous" refers to the fact although in real life the {{{x}}} and {{{y}}} values jump from one integer to another, a function/equation used "as a continuous model" applies to any real value of {{{x}}} , yields real values of {{{y}}} , and has a graph that is a curve flowing continuously through an infinite number of points, without any gaps.
With that understanding, I would say that {{{y=12053*1.0345^x}}} is one equation for the population of the town as a continuous model, but I believe that is not the expected answer.
For any number, equation, or function, there is a variety of representations limited only by your imagination. An exponential function can be written with any base, but in higher math the irrational number
{{{e=approximately}}}{{{2.71828}}} is a very popular base.
Using base {{{e}}} logarithms of both sides of the equal sign in {{{y=12053*1.0345^x}}} ,
we can work our to an equivalent expression:
{{{y=12053*1.0345^x}}}
{{{ln(y)=ln(12053*1.0345^x)}}}
{{{ln(y)=ln(12053)+ln(1.0345^x)}}}
{{{ln(y)=ln(12053)+x*ln(1.0345)}}}
{{{e^ln(y)=e^(ln(12053)+x*ln(1.0345))}}}
{{{e^ln(y)=e^ln(12053)*e^(x*ln(1.0345))}}}
{{{y=12053*e^(x*ln(1.0345))}}}
 
3. Of course the population is increasing.
On the year 2000, {{{x=0}}} and the population is {{{y=12053}}} .
One year later, the model predicts that for {{{x=1}}} that number will have multiplied times {{{1.0345}}} .
That is {{{0.0345}}} times more.
It is {{{0.0345=3.45/100="3.45%"}}} more in one year.
The population is increasing by {{{highlight("3.45%")}}} per year.
You may be expected to show by calculating the relative change
from {{{y(0)=12053}}} for {{{x=0}}}
to {{{y=12053*1.0345}}} for {{{x=1}}}
as {{{(12053*1.0345-12053)/12053=12053*(1.0345-1)/12053=0.0345="3.45%"}}} ,
or you may be expected to apply a formula such as
{{{r-1}}} to find the percent increase as a decimal for {{{P=P[0]*e^rt}}} .