VCE Maths Methods Logs & Exponentials Mini Test 2

Number of marks: 8

Reading time: 2 minutes

Writing time: 12 minutes

Section A Calculator Allowed
Instructions
• Answer all questions in pencil on your Multiple-Choice Answer Sheet.
• Choose the response that is correct for the question.
• A correct answer scores 1; an incorrect answer scores 0.
• Marks will not be deducted for incorrect answers.
• No marks will be given if more than one answer is completed for any question.
• Unless otherwise indicated, the diagrams in this book are not drawn to scale.


Question 1 [2019 Exam 2 Section A Q20]

The expression \(\log_x(y) + \log_y(z)\), where \(x, y\) and \(z\) are all real numbers greater than 1, is equal to

  • A. \(-\frac{1}{\log_y(x)} - \frac{1}{\log_z(y)}\)
  • B. \(\frac{1}{\log_x(y)} + \frac{1}{\log_y(z)}\)
  • C. \(-\frac{1}{\log_x(y)} - \frac{1}{\log_y(z)}\)
  • D. \(\frac{1}{\log_y(x)} + \frac{1}{\log_z(y)}\)
  • E. \(\log_y(x) + \log_z(y)\)
Correct Answer: D
Click here for full solution
Question 2 [2017 Exam 2 Section A Q8]

If \(y = a^{b-4x} + 2\), where \(a > 0\), then \(x\) is equal to

  • A. \(\frac{1}{4}(b - \log_a(y-2))\)
  • B. \(\frac{1}{4}(b - \log_a(y+2))\)
  • C. \(b - \log_a(\frac{1}{4}(y+2))\)
  • D. \(\frac{b}{4} - \log_a(y-2)\)
  • E. \(\frac{1}{4}(b+2 - \log_a(y))\)
Correct Answer: A
Click here for full solution

End of Section A


Section B – No Calculator
Instructions
• Answer all questions in the spaces provided.
• Write your responses in English.
• In questions where a numerical answer is required, an exact value must be given unless otherwise specified.
• In questions where more than one mark is available, appropriate working must be shown.
• Unless otherwise indicated, the diagrams in this book are not drawn to scale.


Question 1 [2023 Exam 1 Q2]

Solve \( e^{2x} - 12 = 4e^x \) for \( x \in \mathbb{R} \). 3 marks

Question 2 [2020 Exam 1 Q4]

Solve the equation \(2\log_2(x+5) - \log_2(x+9) = 1\). 3 marks


End of examination questions

VCE is a registered trademark of the VCAA. The VCAA does not endorse or make any warranties regarding this study resource. Past VCE exams and related content can be accessed directly at www.vcaa.vic.edu.au

>