Given that ${2^m} = 8$ and ${2^n} = 16$, write down the value of $m$ and of $n$.

Hence or otherwise solve ${8^{2x + 1}} = {16^{2x - 3}}$.

$m = 3,{\text{ }}n = 4$     A1A1     N2

[2 marks]

attempt to apply ${({2^a})^b} = {2^{ab}}$     (M1)

eg$\;\;\;6x + 3,{\text{ }}4(2x - 3)$

equating their powers of $2$ (seen anywhere)     M1

eg$\;\;\;3(2x + 1) = 8x - 12$

correct working     A1

eg$\;\;\;8x - 12 = 6x + 3,{\text{ }}2x = 15$

$x = \frac{{15}}{2}\;\;\;(7.5)$     A1     N2

[4 marks]

Total [6 marks]

Indices laws were well understood with many candidates solving the equation correctly. Some candidates used logs, which took longer, and errors crept in.

Indices laws were well understood with many candidates solving the equation correctly. Some candidates used logs, which took longer, and errors crept in.