Expressing a Logarithm in Terms of Given Logs

These questions test the log laws in reverse: you are given a couple of logarithms and must build a new one from them. The mathematical work is small; the insight is rewriting the target number using only the bases you have, here $2$ and $3$.

Given that $\log_a 2 = p$ and $\log_a 3 = q$, express $\log_a 4.5$ in terms of $p$ and $q$. [3]

The working

Step 1, rewrite $4.5$ using $2$ and $3$ only. Note $4.5 = \dfrac{9}{2} = \dfrac{3^2}{2}$: $\log_a 4.5 = \log_a\!\left(\frac{3^2}{2}\right) \quad \text{(M1)}$

Step 2, split with the quotient and power laws ($\log\frac{m}{n} = \log m - \log n$, and $\log m^k = k\log m$): $= \log_a 3^2 - \log_a 2 = 2\log_a 3 - \log_a 2 \quad \text{(M1)}$

Step 3, substitute the given values $\log_a 3 = q$ and $\log_a 2 = p$: $= 2q - p \quad \text{(A1)}$

Where the marks are won and lost

• The decomposition $4.5 = \frac{3^2}{2}$ is the whole question. Any correct route works ($4.5 = \frac{9}{2}$, or $9 \times 2^{-1}$), but it must use only $2$s and $3$s.
• The power law turns $\log_a 3^2$ into $2\log_a 3 = 2q$. Forgetting to bring the index down gives $q$ instead of $2q$.
• The quotient becomes a subtraction, not a division of the logs. $\log\frac{9}{2} = \log 9 - \log 2$, never $\frac{\log 9}{\log 2}$.

Common mistakes

• Writing $\log_a 4.5 = \frac{\log_a 9}{\log_a 2}$ (confusing the quotient law with change of base).
• Dropping the power: $\log_a 3^2 = q$ instead of $2q$.
• Trying to find a numerical value, the answer is meant to stay in terms of $p$ and $q$.

Full method: Laws of Logarithms notes. Topic home: Logs & Exponentials pillar.