The population of Adamsville grew from 10,000 to 16,000 in 6 years. Assuming uninhibited exponential growth,what is the expected population in an additional 4 years?

[tex]\begin{gathered} A=P(1+r)^t \\ \text{Where A = final population in at a time }=\text{ 16,000} \\ P\text{ = initial population }=\text{ 10,000} \\ r=\text{ population growth rate =?} \\ t=\text{ time in years }=\text{ 6} \end{gathered}[/tex]

So, before we can find the expected population in the next 4 years, we need to find the population growth rate r,

Now substituting the values above into the equation we have

[tex]\begin{gathered} A=P(1+r)^t \\ 16,000=10,000(1+r)^6 \\ (1+r)^6=\frac{16,000}{10,000} \\ (1+r)^6=1.6 \\ \text{taking the 6th root of both sides } \\ \sqrt[6]{(1+r)^6}=\sqrt[6]{1.6} \\ 1+r=1.081484 \\ r=1.081484-1 \\ r=0.081484 \end{gathered}[/tex]

So, we have found the rate, which is about 8.15%,

Now, the expected population in 4 years becomes

[tex]\begin{gathered} A=P(1+r)^t \\ the\text{ current population will be 16,000, } \\ \text{time = 4 years } \\ r\text{ = 0.081484} \end{gathered}[/tex]

Substituting the values, we have

[tex]\begin{gathered} A=P(1+r)^t \\ A=16,000(1+0.081484)^4 \\ A=16,000(1.081484)^4 \\ A=21,887.69212 \\ A=21,888\text{ to the nearest whole number } \end{gathered}[/tex]

Hence, the answer is 21,888 to the nearest whole number