Worked Solutions
Exponential & Logarithmic Functions — Worked Solutions (Preliminary Maths Advanced)
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Worked examples for Preliminary Maths Advanced exponential and logarithmic functions. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Logarithm laws
Question
Express $2\log_3 6 - \log_3 4$ as a single logarithm and hence evaluate it.
Solution
Move the coefficient inside first using the power law: $2\log_3 6 = \log_3 6^2 = \log_3 36$.
Now apply the quotient law: $\log_3 36 - \log_3 4 = \log_3 \frac{36}{4} = \log_3 9$.
Since $9 = 3^2$, $\log_3 9 = 2$.
Answer $2$. Deal with the coefficient before combining — that's the step people skip.
Three log laws are in play, so let's take them in order. The coefficient $2$ in front of $\log_3 6$ becomes a power: by the power law, $2\log_3 6 = \log_3 6^2 = \log_3 36$.
Now we have $\log_3 36 - \log_3 4$. A subtraction of logs (same base) becomes a division inside — the quotient law: $\log_3 \frac{36}{4} = \log_3 9$.
Finally, evaluate. We ask "$3$ to what power gives $9$?" — and $3^2 = 9$, so $\log_3 9 = 2$.
The value is $2$. The pattern to remember: coefficients turn into powers, sums into products, differences into quotients.
Apply the log laws in turn.
- Power law: $2\log_3 6 = \log_3 36$
- Quotient law: $\log_3 36 - \log_3 4 = \log_3 \frac{36}{4} = \log_3 9$
- $9 = 3^2 \Rightarrow \log_3 9 = 2$
Value: $2$.
Where the marks go
- 1 mark: Uses the power law: $2\log_3 6 = \log_3 36$
- 1 mark: Uses the quotient law to reach $\log_3 9$
- 1 mark: Evaluates $\log_3 9 = 2$
Key idea
Coefficients become powers, sums become products and differences become quotients — combine to a single log, then evaluate from the base.
Example 2 — Solving an exponential equation
Question
Solve $5^{2x} = 20$, giving your answer correct to three decimal places.
Solution
Take logs of both sides: $\log 5^{2x} = \log 20$.
Bring the power down: $2x \log 5 = \log 20$, so $2x = \dfrac{\log 20}{\log 5}$.
Then $x = \dfrac{\log 20}{2\log 5} = \dfrac{1.30103}{2(0.69897)} = \dfrac{1.30103}{1.39794} \approx 0.931$.
$x \approx 0.931$. Keep full accuracy until the final rounding.
The unknown is stuck in the exponent, and logs are the tool that brings it down. Take the log of both sides: $\log 5^{2x} = \log 20$.
By the power law, the exponent comes out the front: $2x \log 5 = \log 20$. Now it's just a linear equation in $x$.
Divide to isolate $x$: $x = \dfrac{\log 20}{2\log 5}$. Putting the numbers in, $\log 20 \approx 1.30103$ and $\log 5 \approx 0.69897$, so $x = \dfrac{1.30103}{1.39794} \approx 0.9307$.
Rounded to three decimal places, $x \approx 0.931$. The big idea: whenever the variable is an exponent, take logs to pull it down to ground level.
Take logs; bring down the power.
- $\log 5^{2x} = \log 20$
- $2x \log 5 = \log 20$
- $x = \dfrac{\log 20}{2\log 5}$
- $= \dfrac{1.30103}{1.39794} \approx 0.9307$
$x \approx 0.931$.
Where the marks go
- 1 mark: Takes logs of both sides
- 1 mark: Brings the power down: $2x\log 5 = \log 20$
- 1 mark: Correct answer $x \approx 0.931$
Key idea
When the unknown is in the exponent, take logs of both sides and use the power law to bring it down, then solve the resulting linear equation.