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.
In short: full step-by-step worked solutions for Preliminary Maths Advanced — Exponential & Logarithmic Functions. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
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.
Frequently asked questions
Step-by-step solutions to Exponential & Logarithmic Functions questions in Preliminary Maths Advanced, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Advanced, using the methods and notation expected in exams and assessments.