1. **Problem statement:** We have a medicine whose amount decreases exponentially in the body with a half-life $T_{1/2} = 7$ hours. Initially, at $x=0$, the amount is $b=50$ mg. 2. **Formula and explanation:** The exponential decay model is given by: $$f(x) = b \cdot a^x$$ where $a$ is the decay factor per hour. The half-life means that after 7 hours, the amount is half the initial: $$f(7) = b \cdot a^7 = \frac{b}{2}$$ 3. **Find $a$:** Divide both sides by $b$: $$a^7 = \frac{1}{2}$$ Taking the 7th root: $$a = \sqrt[7]{\frac{1}{2}} = 2^{-\frac{1}{7}}$$ 4. **Calculate the percentage decrease per hour:** The percentage decrease per hour is: $$100\% \times (1 - a) = 100\% \times \left(1 - 2^{-\frac{1}{7}}\right)$$ 5. **Model function:** $$f(x) = 50 \cdot \left(2^{-\frac{1}{7}}\right)^x = 50 \cdot 2^{-\frac{x}{7}}$$ 6. **Percentage remaining after 12 hours:** $$f(12) = 50 \cdot 2^{-\frac{12}{7}}$$ Percentage of original amount: $$\frac{f(12)}{50} \times 100\% = 100\% \cdot 2^{-\frac{12}{7}}$$ 7. **Time when amount is 2 mg (no longer effective):** Set: $$50 \cdot 2^{-\frac{x}{7}} = 2$$ Divide both sides by 50: $$2^{-\frac{x}{7}} = \frac{2}{50} = \frac{1}{25}$$ Take logarithm base 2: $$-\frac{x}{7} = \log_2 \left(\frac{1}{25}\right) = -\log_2 25$$ Multiply both sides by $-7$: $$x = 7 \log_2 25$$ Calculate: $$\log_2 25 = \frac{\ln 25}{\ln 2} \approx \frac{3.2189}{0.6931} \approx 4.6439$$ So: $$x \approx 7 \times 4.6439 = 32.5073 \text{ hours}$$ 8. **Interpretation:** The medicine is no longer effective after approximately 32.5 hours, which should match the graph showing the amount dropping below 2 mg around this time.