Question
Let \({\log _3}p = 6\) and \({\log _3}q = 7\) .
Find \({\log _3}{p^2}\) .
[2]
a.
Find \({\log _3}\left( {\frac{p}{q}} \right)\) .
[2]
b.
Find \({\log _3}(9p)\) .
[3]
c.
Markscheme
METHOD 1
evidence of correct formula (M1)
eg \(\log {u^n} = n\log u\) , \(2{\log _3}p\)
\({\log _3}({p^2}) = 12\) A1 N2
METHOD 2
valid method using \(p = {3^6}\) (M1)
eg \({\log _3}{({3^6})^2}\) , \(\log {3^{12}}\) , \(12{\log _3}3\)
\({\log _3}({p^2}) = 12\) A1 N2
[2 marks]
a.
METHOD 1
evidence of correct formula (M1)
eg \(\log \left( {\frac{p}{q}} \right) = \log p - \log q\) , \(6 - 7\)
\({\log _3}\left( {\frac{p}{q}} \right) = - 1\) A1 N2
METHOD 2
valid method using \(p = {3^6}\) and \(q = {3^7}\) (M1)
eg \({\log _3}\left( {\frac{{{3^6}}}{{{3^7}}}} \right)\) , \(\log {3^{ - 1}}\) , \( - {\log _3}3\)
\({\log _3}\left( {\frac{p}{q}} \right) = - 1\) A1 N2
[2 marks]
b.
METHOD 1
evidence of correct formula (M1)
eg \({\log _3}uv = {\log _3}u + {\log _3}v\) , \(\log 9 + \log p\)
\({\log _3}9 = 2\) (may be seen in expression) A1
eg \(2 + \log p\)
\({\log _3}(9p) = 8\) A1 N2
METHOD 2
valid method using \(p = {3^6}\) (M1)
eg \({\log _3}(9 \times {3^6})\) , \({\log _3}({3^2} \times {3^6})\)
correct working A1
eg \({\log _3}9 + {\log _3}{3^6}\) , \({\log _3}{3^8}\)
\({\log _3}(9p) = 8\) A1 N2
[3 marks]
Total [7 marks]
c.
Examiners report
This question proved to be surprisingly challenging for many candidates. A common misunderstanding was to set \(p\) equal to \(6\) and \(q\) equal to \(7\). A large number of candidates had trouble applying the rules of logarithms, and made multiple errors in each part of the question. Common types of errors included incorrect working such as \({\log _3}{p^2} = 36\) in part (a), \({\log _3}\left( {\frac{p}{q}} \right) = \frac{{{{\log }_3}6}}{{{{\log }_3}7}}\) or \({\log _3}\left( {\frac{p}{q}} \right) = {\log _3}6 - {\log _3}7\) in part (b), and \({\log _3}(9p) = 54\) in part (c).
a.
This question proved to be surprisingly challenging for many candidates. A common misunderstanding was to set \(p\) equal to \(6\) and \(q\) equal to \(7\). A large number of candidates had trouble applying the rules of logarithms, and made multiple errors in each part of the question. Common types of errors included incorrect working such as \({\log _3}{p^2} = 36\) in part (a), \({\log _3}\left( {\frac{p}{q}} \right) = \frac{{{{\log }_3}6}}{{{{\log }_3}7}}\) or \({\log _3}\left( {\frac{p}{q}} \right) = {\log _3}6 - {\log _3}7\) in part (b), and \({\log _3}(9p) = 54\) in part (c).
b.
This question proved to be surprisingly challenging for many candidates. A common misunderstanding was to set \(p\) equal to \(6\) and \(q\) equal to \(7\). A large number of candidates had trouble applying the rules of logarithms, and made multiple errors in each part of the question. Common types of errors included incorrect working such as \({\log _3}{p^2} = 36\) in part (a), \({\log _3}\left( {\frac{p}{q}} \right) = \frac{{{{\log }_3}6}}{{{{\log }_3}7}}\) or \({\log _3}\left( {\frac{p}{q}} \right) = {\log _3}6 - {\log _3}7\) in part (b), and \({\log _3}(9p) = 54\) in part (c).
c.
Syllabus sections
