Science:Math Exam Resources/Courses/MATH103/April 2016/Question 06 (a) (v)/Solution 1

From parts (ii) and (iv), we have that if $y={\frac {N}{K}}$ and $N(0)=5\times 10^{6},$ then ${\frac {y}{1-y}}={\frac {1}{4}}e^{\frac {t}{2}}.$

Let $t_{0}$ be the time at which the population reaches 50% of the carrying capacity.

Then, at time $t_{0},$ we know that $N(t_{0})={\frac {1}{2}}K,$ and therefore $y(t_{0})={\frac {N(t_{0})}{K}}={\frac {1}{2}}.$

Thus, we solve the following equation to find $t_{0}$:

${\frac {1/2}{1-1/2}}={\frac {1}{4}}e^{t_{0}/2}.$

So

$e^{t_{0}/2}=4\implies t_{0}=2\ln(4)=\color {blue}\ln(16).$