Let the oak tree population be represented as y.
The rate of population decrease = 8% per year
Since the population continues to decrease at a constant rate, the population size of the oak tree at the end of each year is evaluated as
[tex]\begin{gathered} P(t)=P_0(1-r)^t\text{ ------ equation 1} \\ \text{where} \\ P_0\Rightarrow the\text{ inital poplulation size of the oak trees} \\ t\Rightarrow period,\text{ in years} \\ r\Rightarrow\text{ rate of population decrease} \\ \\ \end{gathered}[/tex]
Thus, when the population reaches half its original value,
[tex]\begin{gathered} P(t)_{}=\frac{y}{2}\text{ (half the or}iginal\text{ value)} \\ P_0=y\text{ (original value)} \\ r\text{ = 0.08 } \\ t\text{ is the period, which is unknown.} \end{gathered}[/tex]
Substitute the above values in equation 1
[tex]\begin{gathered} \frac{y}{2}=y(1-0.08)^t \\ \frac{y}{2}=y\times0.92^t \\ \Rightarrow\frac{1}{2}=0.92^t \\ \end{gathered}[/tex]
Take the logarithm of both sides
[tex]\begin{gathered} \log _{}(\frac{1}{2})=log(0.92^t) \\ \Rightarrow\log _{}(\frac{1}{2})=t\times log(0.92^{}) \\ -0.30=t\times-0.036 \\ -0.30=-0.036t \end{gathered}[/tex]
Divide both sides by the coefficient of t
[tex]\begin{gathered} \text{the co}efficient\text{ of t is -0.036.} \\ \text{thus,} \\ \frac{-0.30}{-0.036}=\frac{-0.036t}{-0.036} \\ \Rightarrow t=8.33 \end{gathered}[/tex]
Hence, the population will reach half of its original value in approximately 8.3 years (nearest tenth).
