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

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

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

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

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

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

So

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