Answer
\(N(t)=3780000(1.027)^t\)
Explanation
To solve this, we are going to use the exponential growth function:
\(f(t)=a(1+b)^t\)
where
\(f(t)\) is the final population after \(t\) years of growth
\(a\) is the initial population
\(b\) is the growth rate in decimal form
\(t\) is the time in years
In our model \(t\) will be the number of years after 2004, so the number of years after 2004 is 0 and the number of years after 2009 is 5. The initial population is, therefore, the number of licensed drivers in 2004, so \(a=3.78million=3780000\). The final population is the number of licensed drivers in 2009, so \(f(t)=4.32million=4320000\). Let's use those facts to calculate the growth rate and complete our model:
For the final population
\(t=5\), \(a=3780000\), and \(f(t)=4320000\). Let's replace the values in our function
\(f(t)=a(1+b)^t\)
\(4320000=3780000(1+b)^5\)
\(\frac{4320000}{3780000} =(1+b)^5\)
\(1+b=\sqrt[5]{\frac{4320000}{3780000}}\)
\(b=\sqrt[5]{\frac{4320000}{3780000}}-1\)
\(b=0.027\)
Now we can complete our model
\(f(t)=a(1+b)^t\)
\(N(t)=3780000(1+0.027)^t\)
\(N(t)=3780000(1.027)^t\)
where \(t\) is the number of years after 2004
