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Question
Find \({\log _2}32\) .
Given that \({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right)\) can be written as \(px + qy\) , find the value of p and of q.
Answer/Explanation
Markscheme
5 A1 N1
[1 mark]
METHOD 1
\({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right) = {\log _2}{32^x} – {\log _2}{8^y}\) (A1)
\( = x{\log _2}32 – y{\log _2}8\) (A1)
\({\log _2}8 = 3\) (A1)
\(p = 5\) , \(q = – 3\) (accept \(5x – 3y\) ) A1 N3
METHOD 2
\(\frac{{{{32}^x}}}{{{8^y}}} = \frac{{{{({2^5})}^x}}}{{{{({2^3})}^y}}}\) (A1)
\( = \frac{{{2^5}^x}}{{{2^3}^y}}\) (A1)
\( = {2^{5x – 3y}}\) (A1)
\({\log _2}({2^{5x – 3y}}) = 5x – 3y\)
\(p = 5\) , \(q = – 3\) (accept \(5x – 3y\) ) A1 N3
[4 marks]
Question
Solve \({\log _2}x + {\log _2}(x – 2) = 3\) , for \(x > 2\) .
Answer/Explanation
Markscheme
recognizing \(\log a + \log b = \log ab\) (seen anywhere) (A1)
e.g. \({\log _2}(x(x – 2))\) , \({x^2} – 2x\)
recognizing \({\log _a}b = x \Leftrightarrow {a^x} = b\) (A1)
e.g. \({2^3} = 8\)
correct simplification A1
e.g. \(x(x – 2) = {2^3}\) , \({x^2} – 2x – 8\)
evidence of correct approach to solve (M1)
e.g. factorizing, quadratic formula
correct working A1
e.g. \((x – 4)(x + 2)\) , \(\frac{{2 \pm \sqrt {36} }}{2}\)
\(x = 4\) A2 N3
[7 marks]
Question
Write down the value of
(i) \({\log _3}27\);
(ii) \({\log _8}\frac{1}{8}\);
(iii) \({\log _{16}}4\).
Hence, solve \({\log _3}27 + {\log _8}\frac{1}{8} – {\log _{16}}4 = {\log _4}x\).
Answer/Explanation
Markscheme
(i) \({\log _3}27 = 3\) A1 N1
[1 mark]
(ii) \({\log _8}\frac{1}{8} = – 1\) A1 N1
[1 mark]
(iii) \({\log _{16}}4 = \frac{1}{2}\) A1 N1
[1 mark]
correct equation with their three values (A1)
eg \(\frac{3}{2} = {\log _4}x{\text{, }}3 + ( – 1) – \frac{1}{2} = {\log _4}x\)
correct working involving powers (A1)
eg \(x = {4^{\frac{3}{2}}}{\text{, }}{4^{\frac{3}{2}}} = {4^{{{\log }_4}x}}\)
\(x = 8\) A1 N2
[3 marks]
Question
Find the value of each of the following, giving your answer as an integer.
\({\log _6}36\)
\({\log _6}4 + {\log _6}9\)
\({\log _6}2 – {\log _6}12\)
Answer/Explanation
Markscheme
correct approach (A1)
eg \({6^x} = 36,{\text{ }}{6^2}\)
\(2\) A1 N2
[2 marks]
correct simplification (A1)
eg \({\log _6}36,{\text{ }}\log (4 \times 9)\)
\(2\) A1 N2
[2 marks]
correct simplification (A1)
eg \({\log _6}\frac{2}{{12}},{\text{ }}\log (2 \div 12)\)
correct working (A1)
eg \({\log _6}\frac{1}{6},{\text{ }}{6^{ – 1}} = \frac{1}{6}{,6^x} = \frac{1}{6}\)
\(-1\) A1 N2
[3 marks]
Question
Let \(x = \ln 3\) and \(y = \ln 5\). Write the following expressions in terms of \(x\) and \(y\).
\(\ln \left( {\frac{5}{3}} \right)\).
\(\ln 45\).
Answer/Explanation
Markscheme
correct approach (A1)
eg\(\,\,\,\,\,\)\(\ln 5 – \ln 3\)
\(\ln \left( {\frac{5}{3}} \right) = y – x\) A1 N2
[2 marks]
recognizing factors of 45 (may be seen in log expansion) (M1)
eg\(\,\,\,\,\,\)\(\ln (9 \times 5),{\text{ }}3 \times 3 \times 5,{\text{ }}\log {3^2} \times \log 5\)
correct application of \(\log (ab) = \log a + \log b\) (A1)
eg\(\,\,\,\,\,\)\(\ln 9 + \ln 5,{\text{ }}\ln 3 + \ln 3 + \ln 5,{\text{ }}\ln {3^2} + \ln 5\)
correct working (A1)
eg\(\,\,\,\,\,\)\(2\ln 3 + \ln 5,{\text{ }}x + x + y\)
\(\ln 45 = 2x + y\) A1 N3
[4 marks]
Question
An arithmetic sequence has \({u_1} = {\text{lo}}{{\text{g}}_c}\left( p \right)\) and \({u_2} = {\text{lo}}{{\text{g}}_c}\left( {pq} \right)\), where \(c > 1\) and \(p,\,\,q > 0\).
Show that \(d = {\text{lo}}{{\text{g}}_c}\left( q \right)\).
Let \(p = {c^2}\) and \(q = {c^3}\). Find the value of \(\sum\limits_{n = 1}^{20} {{u_n}} \).
Answer/Explanation
Markscheme
valid approach involving addition or subtraction M1
eg \({u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} – {u_2}\)
correct application of log law A1
eg \({\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)\)
\(d = {\text{lo}}{{\text{g}}_c}\,q\) AG N0
[2 marks]
METHOD 1 (finding \({u_1}\) and d)
recognizing \(\sum { = {S_{20}}} \) (seen anywhere) (A1)
attempt to find \({u_1}\) or d using \({\text{lo}}{{\text{g}}_c}\,{c^k} = k\) (M1)
eg \({\text{lo}}{{\text{g}}_c}\,c\), \({\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c\), correct value of \({u_1}\) or d
\({u_1}\) = 2, d = 3 (seen anywhere) (A1)(A1)
correct working (A1)
eg \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)
\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610 A1 N2
METHOD 2 (expressing S in terms of c)
recognizing \(\sum { = {S_{20}}} \) (seen anywhere) (A1)
correct expression for S in terms of c (A1)
eg \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)
\({\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3\) (seen anywhere) (A1)(A1)
correct working (A1)
eg \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)
\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610 A1 N2
METHOD 3 (expressing S in terms of c)
recognizing \(\sum { = {S_{20}}} \) (seen anywhere) (A1)
correct expression for S in terms of c (A1)
eg \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)
correct application of log law (A1)
eg \(2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)\)
correct application of definition of log (A1)
eg \({\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57\)
correct working (A1)
eg \({S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)\)
\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610 A1 N2
[6 marks]