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Date May 2013 Marks available 2 Reference code 13M.1.sl.TZ2.3
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 3 Adapted from N/A

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

Topic 1 - Algebra » 1.2 » Laws of exponents; laws of logarithms.
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