IB DP Math MAA SL : IB Style Mock Exams – Set 5 Paper 1

(a)
\(\log_{10} 24 = \log_{10} (2^3 \times 3)\)
Using log laws: \(= \log_{10} (2^3) + \log_{10} 3\)
\(= 3\log_{10} 2 + \log_{10} 3\)
Substituting \(p\) and \(q\):
\(= 3p + q\)

(b)
Using the change of base formula:
\(\log_{3} 8 = \frac{\log_{10} 8}{\log_{10} 3}\)
\(= \frac{\log_{10} (2^3)}{q}\)
\(= \frac{3\log_{10} 2}{q}\)
\(= \frac{3p}{q}\)