(i)     ${\log _3}27$;

(ii)     ${\log _8}\frac{1}{8}$;

(iii)     ${\log _{16}}4$.

Hence, solve ${\log _3}27 + {\log _8}\frac{1}{8} - {\log _{16}}4 = {\log _4}x$.

(i)     ${\log _3}27 = 3$     A1     N1

[1 mark]

(ii)     ${\log _8}\frac{1}{8} =  - 1$     A1     N1

[1 mark]

(iii)     ${\log _{16}}4 = \frac{1}{2}$     A1     N1

[1 mark]

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]