## Question 5(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_{10} n = \log_{10} a + kt\log_{10} 2$ | M1 | AO 1.1a | AG Allow $t\log 2^k$ |
| This is of form $y = mx + c$ [with $\log_{10} n$ as $y$ and $t$ as $x$] | E1 [2] | AO 1.1 | |

## Question 5(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Reasonable line of best fit drawn (by eye) | B1 | AO 1.1a | With $0.9 < c < 1.2$ |
| Suitable method leading to $a$ value e.g. use of intercept leading to $0.9 < \log a < 1.2$, so $7.4 < a < 15.85$ | M1 | AO 2.2a | May use 2 points from line or condone use of 2 given points |
| Suitable method leading to $k$ value e.g. $k\log_{10} 2 =$ gradient $\approx 0.33$ | M1 | AO 1.1 | Finding gradient of line or sub'n of $t$ and $\log n$ |
| $k$ in range $0 < k < 1.25$ and $a$ in range $7.4 < a < 15.85$ | A1 [4] | AO 2.2a | If gradient of exactly 1/3 used $k = 1.10730936...$ |

## Question 5(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $500000 = 10 \times 2^{1.1t}$ | M1 | AO 3.4 | Correct substitution |
| $1.1t\log 2 = \log 50000$ | | | |
| $t = 14.2$ | A1 | AO 1.1 | Value of $t$ (FT their $a$ and $k$); For $k=1.10730936...$, $t=14.1$, same answer |
| $t = 14$ is $1/3/18$ | M1 | AO 3.4 | Translation into date |
| So $1/4/18$ | A1 [4] | AO 3.2a | Rounding up |

## Question 5(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Suitable reason e.g. the data are only for a short time scale and cannot extrapolate; e.g. there will not be enough people for the growth to continue | E1 [1] | AO 3.5b | |