| Date | May 2014 | Marks available | 3 | Reference code | 14M.1.sl.TZ1.4 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Hence and Solve | Question number | 4 | Adapted from | N/A |
Question
Write down the value of
(i) \({\log _3}27\);
[1]
a(i).
(ii) \({\log _8}\frac{1}{8}\);
[1]
a(ii).
(iii) \({\log _{16}}4\).
[1]
a(iii).
Hence, solve \({\log _3}27 + {\log _8}\frac{1}{8} - {\log _{16}}4 = {\log _4}x\).
[3]
b.
Markscheme
(i) \({\log _3}27 = 3\) A1 N1
[1 mark]
a(i).
(ii) \({\log _8}\frac{1}{8} = - 1\) A1 N1
[1 mark]
a(ii).
(iii) \({\log _{16}}4 = \frac{1}{2}\) A1 N1
[1 mark]
a(iii).
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]
b.
Examiners report
[N/A]
a(i).
[N/A]
a(ii).
[N/A]
a(iii).
[N/A]
b.
