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Date	May 2017	Marks available	4	Reference code	17M.1.AHL.TZ1.H_1
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	H_1	Adapted from	N/A

Question

Find the solution of ${log}_{2} x - {log}_{2} 5 = 2 + {log}_{2} 3$ ${\log _2}x - {\log _2}5 = 2 + {\log _2}3$ .

Markscheme

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

${log}_{2} x - {log}_{2} 5 = 2 + {log}_{2} 3$ ${\log _2}x - {\log _2}5 = 2 + {\log _2}3$

collecting at least two log terms (M1)

eg $\,\,\,\,\,$ ${log}_{2} \frac{x}{5} = 2 + {log}_{2} 3 or {log}_{2} \frac{x}{15} = 2$ ${\log _2}\frac{x}{5} = 2 + {\log _2}3{\text{ or }}{\log _2}\frac{x}{{15}} = 2$

obtaining a correct equation without logs (M1)

eg $\,\,\,\,\,$ $\frac{x}{5} = 12$ $\frac{x}{5} = 12$ $\,\,\,$ OR $\,\,\,$ $\frac{x}{15} = 2^{2}$ $\frac{x}{{15}} = {2^2}$ (A1)

$x = 60$ $x = 60$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.5—Intro to logs

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