Q1
Answer
The value of a car, $V$ pounds, is modelled by the equation $V = ab^t$, where $a$ and $b$ are constants and $t$ is the number of years since the car was purchased. A line, $l$, of $\log_4V$ against $t$ is a straight line which meets the vertical axis at $(0, \log_440000)$ has a gradient of $-0.1$
  1. Write down an equation for $l$
  2. Find, in exact form, the values of $a$ and $b$
  3. With reference to the model, interpret the values of the constants $a$ and $b$
  4. Find the value of the car after 7 years
  5. After how many years is the value of the car less than £10000?
  6. State a limitation of the model.
  1. $\log_4 V = -0.1t + \log_4 40000$
  2. $a = 40000$ and $b = 4^{-0.1}$
  3. $a$ initial value, $b$ ratio of how much it depreciates by per year
  4. $15157.17$
  5. $11$
  6. The model does not include other factors which may affect the value
Q2
Answer
Solve, showing detailed working $$\log_{11}(2x-1)=1-\log_{11}(x+4)$$
$x = \dfrac{3}{2}$
Q3
Answer
The graph of $y = ab^x$ passes through the points $(2, 400)$ and $(5, 50)$.
  1. Find the values of the constants $a$ and $b$
  2. Given that $ab^x < k$, for some constant $k > 0$, show that $x < \dfrac{\log\left(\frac{1600}{k}\right)}{\log2}$, where log means log to any valid base.
  1. $a = 1600$ and $b = \frac{1}{2}$
  2. $\log 1600 + x \log (2)^{-1} < \log k$ and rearrange with log laws