# Question 4:

## Part (a): $\log 5^{x-1} = \log 120$, $(x-1)\log 5 = \log 120$, $x = 3.97$

| Answer/Working | Mark | Guidance |
|---|---|---|
| Introduce logarithms throughout (or $\log_5 120 / \log_{120} 5$) and drop power | M1* | Don't need to see base if taking logs on both sides. If taking logs on one side only, base must be explicit. |
| Obtain $(x-1)\log 5 = \log 120$, or equiv (e.g. $x - 1 = \log_5 120$) | A1 | Condone lack of brackets i.e. $x - 1 \log 5 = \log 120$ as long as clearly implied by later working. |
| Attempt to solve | M1d* | Attempt at correct process i.e. $\frac{\log 120}{\log_5 5} \pm 1$ or equiv. Allow M1 if $\frac{\log 120}{\log_5 5} \pm 1$ subsequently becomes $\log 24 \pm 1$. |
| Obtain $3.97$, or better | A1 (4) | Allow more accurate solution e.g. $3.975$ then isw if rounded to $3.98$. However, $3.98$ without more accurate answer is A0. |

## Part (b): $\log_2 x + \log_2 9 = \log_2(x+5)$, $9x = x+5$, $x = \frac{5}{8}$

| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $2\log 3 = \log 9$ or $\log 3^2$ | B1 | Must be correct statement when done. Condone lack of base throughout. |
| Use $\log a + \log b = \log ab$, or equiv | M1 | Must be used to combine 2 (or more) terms of $\log x + \log k = \log(x+5)$. Could move $\log_2 x$ and/or $\log_2 9$ across to RHS. |
| Obtain correct equation with single log term on each side (or single $\log = 0$) | A1 | e.g. $\log_2(9x) = \log_2(x+5)$. Allow A1 for correct equation with logs removed if several steps run together. |
| Obtain $x = \frac{5}{8}$ | A1 (4) | Allow $0.625$. |