## Question 8:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log 2^{n-3} = \log 18000$ | M1* | Introduce logs and drop power. Can use logs to any base as long as consistent on both sides |
| $(n-3)\log 2 = \log 18000$ | A1 | Or $n-3 = \log_2 18000$. Brackets now need to be seen explicitly |
| $n-3 = 14.1$ | M1d* | Attempt to solve for $n$. Correct order of operations. M0 for $\log_2 18000 - 3$ |
| $n = 17.1$ | A1 | Final answer must be correct for all sig fig shown $(n=17.13570929...)$. 0/4 for answer only or T&I |

# Question (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $2\log_2 x - \log_2 y = 7$ | M1 | Correct use of one log law on a correct equation. Either on first eqn to get $\log_2(xy) = 8$, or on second eqn to get at least $\log_2 x^2 - \log_2 y = 7$. Allow for one correct use, even if error made with other equation. Must be used on a correct equation so M0 if an error has already occurred. |
| $(\log_2 x + \log_2 y) + (2\log_2 x - \log_2 y) = 15$ | M1 | Attempt to eliminate one variable. To get an equation in just one variable, which may or may not still involve logs. Must be a sound algebraic process with the two equations, though errors may have been made earlier with log/index laws. |
| $3\log_2 x = 15$ | A1 | Obtain correct equation in just one variable. Possible equations are $3\log_2 x = 15$, $\log_2 x^3 = 15$, $x^3 = 32768$ or $3\log_2 y = 9$, $\log_2 y^3 = 9$, $y^3 = 512$. |
| $x = 2^5$ | M1 | Correctly use $2^k$ as inverse of $\log_2$. At any stage. M0 for e.g. $\log_2 x + \log_2 y = 8$ becoming $x + y = 2^8$. |
| $x = 32,\ y = 8$ | A1 | Obtain $x = 32,\ y = 8$. Both values required, and no others. Answer only with no evidence of log or index work is 0/5. |
| | **[5]** | |

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