Solution — 2025 Question 6 Exponential & Logarithmic Functions

(a)

\begin{matrix} A = 705 ∎ \end{matrix}

(b)

When $t = 20,$ $N = 282,$

\begin{matrix} 705 e^{- 20 q} & = 282 \\ e^{- 20 q} & = \frac{2}{5} \\ \ln e^{- 20 q} & = \ln \frac{2}{5} \\ - 20 q & = \ln \frac{2}{5} \\ q & = - \frac{\ln \frac{2}{5}}{20} \\ = 0.045 815 \\ = 0.0458 ∎ (to 3 s.f.) \end{matrix}

(c)

When $N = 200,$

\begin{matrix} 705 e^{- 0.045 815 t} & = 200 \\ e^{- 0.045 815 t} & = \frac{40}{141} \\ \ln e^{- 0.045 815 t} & = \ln \frac{40}{141} \\ - 0.045 815 t & = \ln \frac{40}{141} \\ t & = - 21.827 \ln \frac{40}{141} \\ = 27.5 ∎ (to 3 s.f.) \end{matrix}

(d)

Instead of the original model of $N = A e^{- q t},$ where the number of squirrels are decreasing, the new model is $N = A e^{q t},$ since the number of squirrels are increasing.

When $t = 0, N = 282$ $A = 282$

When $t = 10, N = 500$

\begin{matrix} 282 e^{10 q} & = 500 \\ e^{10 q} & = \frac{250}{141} \\ \ln e^{10 q} & = \ln \frac{250}{141} \\ 10 q & = \ln \frac{250}{141} \\ q & = \frac{\ln \frac{250}{141}}{10} \\ = 0.057 270 \end{matrix}

Hence the model that represents exponential growth is

$N = 282 e^{0.057 270 t} ∎$