# Question 8:

## Part (i)(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0, 1)$ | B1 [1] | State $(0, 1)$. Allow no brackets. B1 for $x = 0$, $y = 1$ — must have $x = 0$ stated explicitly. B0 for $y = a^0 = 1$ (as $x = 0$ is implicit) |

## Part (i)(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0, 4)$ | B1 [1] | State $(0, 4)$. Allow no brackets. B1 for $x = 0$, $y = 4$ — must have $x = 0$ stated explicitly. B0 for $y = 4b^0 = 4$ (as $x = 0$ is implicit) |

## Part (i)(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| State a possible value for $a$ | B1 | Must satisfy $a > 1$. Must be a single value. Could be irrational e.g. $e$. Must be fully correct so B0 for e.g. 'any positive number such as 3' |
| State a possible value for $b$ | B1 [2] | Must satisfy $0 < b < 1$. Must be a single value. Could be irrational e.g. $e^{-1}$. Must be fully correct. **SR** allow B1 if both $a$ and $b$ given correctly as a range of values |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_2 a^x = \log_2(4b^x)$ | M1 | Equate $a^x$ and $4b^x$ and introduce logarithms at some stage. Could either use the two given equations, or $b$ could have already been eliminated so using two equations in $a$ only. Must take logs of each side so M0 for $4\log_2(b^x)$. Allow just log, with no base specified, or $\log_2$. Allow logs to any base, or no base, as long as consistent |
| $\log_2 a^x = \log_2 4 + \log_2 b^x$ | M1 | Use $\log ab = \log a + \log b$ correctly. Or correct use of $\log \frac{a}{b} = \log a - \log b$. Used on a correct expression e.g. $\log_2(4b^x)$ or $\log_2 4(\frac{2}{a})^x$. Equation could either have both $a$ and $b$ or just $a$. Must be used on an expression associated with $a^x = 4b^x$, either before or after substitution, so M0 for $\log_2(ab) = 1$, hence $\log_2 a + \log_2 b = 1$ |
| $x\log_2 a = \log_2 4 + x\log_2 b$ | M1 | Use $\log a^b = b\log a$ correctly at least once. Allow if used on an expression that is possibly incorrect. Allow M1 for $x\log_2 a = x\log_2 4b$ as one use is correct. Equation could either have both $a$ and $b$ or just $a$ |
| $x\log_2 a = \log_2 4 + x\log_2\left(\frac{2}{a}\right)$ | B1 | Use $b = \frac{2}{a}$ to produce a correct equation in $a$ and $x$ only. Can be gained at any stage, including before use of logs. If logs involved then allow for no, or incorrect, base as long as equation is fully correct. Could be an equiv method e.g. $(a \times a)^x = 4(a \times b)^x$ hence $a^{2x} = 4 \times 2^x$. Must be eliminating $b$, so $(\frac{2}{b})^x = 4b^x$ is B0 unless the equation is later changed to being in terms of $a$ |
| $x\log_2 a = 2 + x\log_2 2 - x\log_2 a$ $x(2\log_2 a - 1) = 2$ $x = \frac{2}{2\log_2 a - 1}$ **AG** | A1 [5] | Obtain given relationship with no wrong working. Proof must be fully correct with enough detail to be convincing. Must use $\log_2$ throughout proof for A1 — allow 1 slip. Using numerical values for $a$ and $b$ will gain no credit. Working with equations involving $y$ is M0 unless $y$ is subsequently eliminated |