Find the value of ${\log _2}40 - {\log _2}5$ .

Find the value of ${8^{{{\log }_2}5}}$ .

evidence of correct formula    (M1)

eg   $\log a - \log b = \log \frac{a}{b}$ , $\log \left( {\frac{{40}}{5}} \right)$ , $\log 8 + \log 5 - \log 5$

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   ${\log _2}8$ , ${2^3} = 8$

${\log _2}40 - {\log _2}5 = 3$     A1     N2

[3 marks]

attempt to write $8$ as a power of $2$ (seen anywhere)     (M1)

eg   ${({2^3})^{{{\log }_2}5}}$ , ${2^3} = 8$ , ${2^a}$

multiplying powers     (M1)

eg   ${2^{3{{\log }_2}5}}$ , $a{\log _2}5$

correct working     (A1)

eg   ${2^{{{\log }_2}125}}$ , ${\log _2}{5^3}$ , ${\left( {{2^{{{\log }_2}5}}} \right)^3}$

${8^{{{\log }_2}5}} = 125$     A1     N3

[4 marks]

Many candidates readily earned marks in part (a). Some interpreted ${\log _2}40 - {\log _2}5$ to mean $\frac{{{{\log }_2}40}}{{{{\log }_2}5}}$ , an error which led to no further marks. Others left the answer as ${\log _2}5$ where an integer answer is expected.

Part (b) proved challenging for most candidates, with few recognizing that changing $8$ to base $2$ is a helpful move. Some made it as far as ${2^{3{{\log }_2}5}}$ yet could not make that final leap to an integer.