**Method 1** (Substituting $a = 3b$ into second equation at some stage)

Using a law of logs correctly (anywhere) | M1 | e.g. $\log_3 ab = 2$

Substitution of $3b$ for $a$ (or $a/3$ for $b$) | M1 | e.g. $\log_3 3b^2 = 2$

Using base correctly on correctly derived $\log_3 p = q$ | M1 | e.g. $3b^2 = 3^2$

First correct value | A1 | $b = \sqrt{3}$ (allow $3^{1/2}$)

Correct method to find other value (dep. on at least first M mark) | M1 |

Second answer | A1 | $a = 3b = 3\sqrt{3}$ or $\sqrt{27}$

**Method 2** (Working with two equations in $\log_3 a$ and $\log_3 b$)

"Taking logs" of first equation and "separating" | M1 | $\log_3 a = \log_3 3 + \log_3 b$ (= $1 + \log_3 b$)

Solving simultaneous equations to find $\log_3 a$ or $\log_3 b$ | M1 | [$\log_3 a = 1\frac{1}{2}, \log_3 b = \frac{1}{2}$]

Using base correctly to find $a$ or $b$ | M1 |

Correct value for $a$ or $b$ | A1 | $a = 3\sqrt{3}$ or $b = \sqrt{3}$

Correct method for second answer, dep. on first M; correct second answer | M1; A1 | [6]

**Notes:**
- Answers must be exact; decimal answers lose both A marks
- There are several variations on Method 1, depending on the stage at which $a = 3b$ is used, but they should all mark as in scheme.
- In this method, the first three method marks on Epen are for:
  - (i) First M1: correct use of log law,
  - (ii) Second M1: substitution of $a = 3b$,
  - (iii) Third M1: requires using base correctly on correctly derived $\log_3 p = q$

---