17M.2.AHL.TZ2.H_6

pestleMathematicsAAHLPaper 217M· sl-1-5-intro-to-logssource ↗

Given that ${\log _{10}}\left( {\frac{1}{{2\sqrt 2 }}\left( {p + 2q} \right)} \right) = \frac{1}{2}\left( {{{\log }_{10}}p + {{\log }_{10}}q} \right),{\text{ }}p > 0,{\text{ }}q > 0$ , find $p$ in terms of $q$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = \frac{1}{2}({\log _{10}}p + {\log _{10}}q)$

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = \frac{1}{2}{\log _{10}}pq$ (M1)

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = {\log _{10}}{(pq)^{\frac{1}{2}}}$ (M1)

$\frac{1}{{2\sqrt 2 }}(p + 2q) = {(pq)^{\frac{1}{2}}}$ (A1)

${(p + 2q)^2} = 8pq$

${p^2} + 4pq + 4{q^2} = 8pq$

${p^2} - 4pq + 4{q^2} = 0$

${(p - 2q)^2} = 0$ M1

hence $p = 2q$ A1

[5 marks]

Examiners’ report

[N/A]