1. **Problem statement:** A bacteria culture doubles every 4 hours, but an antibiotic kills 50% of the bacteria every 6 hours. We want to find the net growth function of the bacteria over time. 2. **Formulas and rules:** - Doubling every 4 hours means the bacteria population grows by a factor of 2 every 4 hours. - Killing 50% every 6 hours means the population is multiplied by 0.5 every 6 hours. - We model growth and decay using exponential functions. 3. **Growth function:** Let $P_0$ be the initial population. - Growth rate per hour for doubling every 4 hours is $r_g = \frac{\ln(2)}{4}$. - So growth factor after $t$ hours is $e^{r_g t} = e^{\frac{\ln(2)}{4} t} = 2^{\frac{t}{4}}$. 4. **Decay function:** Killing 50% every 6 hours means decay rate per hour is $r_d = \frac{\ln(0.5)}{6}$. - Decay factor after $t$ hours is $e^{r_d t} = e^{\frac{\ln(0.5)}{6} t} = (0.5)^{\frac{t}{6}}$. 5. **Net population function:** Multiply growth and decay factors: $$P(t) = P_0 \times 2^{\frac{t}{4}} \times (0.5)^{\frac{t}{6}}$$ 6. **Simplify the expression:** $$P(t) = P_0 \times 2^{\frac{t}{4}} \times 2^{-\frac{t}{6}} = P_0 \times 2^{\frac{t}{4} - \frac{t}{6}} = P_0 \times 2^{\frac{3t - 2t}{12}} = P_0 \times 2^{\frac{t}{12}}$$ 7. **Interpretation:** The net effect is that the bacteria population grows by a factor of 2 every 12 hours. **Final answer:** $$\boxed{P(t) = P_0 \times 2^{\frac{t}{12}}}$$ This means the bacteria population doubles every 12 hours after considering the antibiotic effect.