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.