Question

sheila is a wildlife biologist. at the beginning of each year, she tracks the wild turkey and white - tail deer population in the game reserve where she works. at the first year sheila counted 12 wild turkeys, and their number increases by approximately 40% each year. at the first year sheila counted 18 white - tail deer, and their number increases by 10 additional deer per year. what is the first year in which sheila counts more turkeys than deer?

Explanation:

Step1: Write the population - growth formulas

The population of wild turkeys forms a geometric - sequence with the first term \(a = 12\) and the common ratio \(r=1 + 0.4=1.4\). So the number of wild turkeys \(T(n)\) in the \(n\)th year is given by the formula \(T(n)=12\times1.4^{n - 1}\). The population of white - tail deer forms an arithmetic sequence with the first term \(a = 18\) and the common difference \(d = 10\). So the number of white - tail deer \(D(n)\) in the \(n\)th year is given by the formula \(D(n)=18+(n - 1)\times10=10n + 8\).

Step2: Set up the inequality

We want to find the first positive integer \(n\) such that \(T(n)>D(n)\), that is \(12\times1.4^{n - 1}>10n + 8\).

Step3: Test values of \(n\)

For \(n = 1\):
\(T(1)=12\) and \(D(1)=18\), \(12<18\).
For \(n = 2\):
\(T(2)=12\times1.4=16.8\) and \(D(2)=18 + 10=28\), \(16.8<28\).
For \(n = 3\):
\(T(3)=12\times1.4^{2}=12\times1.96 = 23.52\) and \(D(3)=18+2\times10 = 38\), \(23.52<38\).
For \(n = 4\):
\(T(4)=12\times1.4^{3}=12\times2.744 = 32.928\) and \(D(4)=18+3\times10 = 48\), \(32.928<48\).
For \(n = 5\):
\(T(5)=12\times1.4^{4}=12\times3.8416 = 46.1952\) and \(D(5)=18+4\times10 = 58\), \(46.1952<58\).
For \(n = 6\):
\(T(6)=12\times1.4^{5}=12\times5.37824 = 64.53888\) and \(D(6)=18+5\times10 = 68\), \(64.53888<68\).
For \(n = 7\):
\(T(7)=12\times1.4^{6}=12\times7.529536=90.354432\) and \(D(7)=18+6\times10 = 78\), \(90.354432>78\).

Answer:

7