1. **Problem statement:** The value of a car $V$ is modeled by $V = ab^t$, where $a$ and $b$ are constants and $t$ is the number of years since purchase. The line $l$ shows the linear relationship between $t$ and $\log_4 V$ with gradient $-\frac{1}{10}$ and intercept $\log_4 40000$. 2. **(a) Equation of line $l$:** The line $l$ relates $t$ and $\log_4 V$ as: $$\log_4 V = -\frac{1}{10}t + \log_4 40000$$ 3. **(b) Find $a$ and $b$:** Given $V = ab^t$, take $\log_4$ of both sides: $$\log_4 V = \log_4 a + t \log_4 b$$ Comparing with line $l$: $$\log_4 a = \log_4 40000 \Rightarrow a = 40000$$ $$\log_4 b = -\frac{1}{10} \Rightarrow b = 4^{-\frac{1}{10}}$$ 4. **(c) Interpretation of $a$ and $b$:** - $a = 40000$ is the initial value of the car at $t=0$ years. - $b = 4^{-\frac{1}{10}}$ is the decay factor per year, showing the car's value decreases by a factor of $b$ each year. 5. **(d) Value after 7 years:** $$V = 40000 \times \left(4^{-\frac{1}{10}}\right)^7 = 40000 \times 4^{-\frac{7}{10}}$$ This is the exact value after 7 years. 6. **(e) Years until value less than 10000:** We want $V < 10000$: $$40000 \times 4^{-\frac{t}{10}} < 10000$$ Divide both sides by 40000: $$4^{-\frac{t}{10}} < \frac{1}{4}$$ Since $\frac{1}{4} = 4^{-1}$, we have: $$4^{-\frac{t}{10}} < 4^{-1}$$ Because the base 4 is greater than 1, inequality reverses when taking logs: $$-\frac{t}{10} < -1 \Rightarrow t > 10$$ So after more than 10 years, the car's value is less than 10000. 7. **(f) Limitation of the model:** The model assumes continuous exponential decay which may not reflect real-world factors like maintenance, market changes, or sudden value drops. Also, it predicts values for all $t \geq 0$ without considering the car's minimum possible value or resale market fluctuations.