Accedmychevron_right11thchevron_rightmathchevron_rightFunctions And Graphschevron_rightExercise 2.2

Solution

Given:

  • Half-life t1/2=2t_{1/2}=2 years
  • Q0=200Q_0=200 grams
  • Q(t)=Q0(12)t/2Q(t)=Q_0\left(\frac{1}{2}\right)^{t/2}

We need Q(6)Q(6):

Q(6)=200(12)6/2=200(12)3=20018=25\begin{aligned} Q(6)&=200\left(\frac{1}{2}\right)^{6/2}\\ &=200\left(\frac{1}{2}\right)^3\\ &=200\cdot\frac{1}{8}\\ &=25 \end{aligned} Q(6)=25 grams\boxed{Q(6)=25\ \text{grams}}

Graphically, this matches the fact that every 2 years the quantity halves:

2001005025(after 6 years)200\to 100\to 50\to 25\quad \text{(after 6 years)}