IBDP Maths AA SL 1.7 Laws of exponents with rational exponents AA HL Paper 1- Exam Style Questions- New Syllabus
Question
Given that \(\log_{10} 2 = p\) and \(\log_{10} 3 = q\), where \(p\) and \(q\) are real numbers, answer the following.
(a) Express \(\log_{10} 24\) in terms of \(p\) and \(q\).
(b) Express \(\log_{3} 8\) in terms of \(p\) and \(q\).
Most-appropriate topic codes (IB Mathematics: Analysis and Approaches HL 2025):
• SL 1.7: Laws of Logarithms & Change of Base — parts (a), (b)
▶️ Answer/Explanation
(a)
We are asked to find \(\log_{10} 24\).
First, factorize 24: \(24 = 8 \times 3 = 2^3 \times 3\).
Using the laws of logarithms:
\(\log_{10} 24 = \log_{10}(2^3 \times 3) = \log_{10}(2^3) + \log_{10} 3\)
\(= 3\log_{10} 2 + \log_{10} 3\)
Given \(\log_{10} 2 = p\) and \(\log_{10} 3 = q\), substitute:
\(\log_{10} 24 = 3p + q\)
\(\boxed{3p + q}\)
(b)
We are asked to find \(\log_{3} 8\).
Use the change of base formula:
\(\log_{3} 8 = \frac{\log_{10} 8}{\log_{10} 3}\)
Express \(\log_{10} 8\) in terms of \(p\): \(\log_{10} 8 = \log_{10}(2^3) = 3\log_{10} 2 = 3p\).
Given \(\log_{10} 3 = q\), substitute:
\(\log_{3} 8 = \frac{3p}{q}\)
\(\boxed{\frac{3p}{q}}\)