$\left\{ {\begin{array}{*{20}{c}}{\log _8^{\frac{{\sqrt 2 }}{4}} = \log _{{2^3}}^{\frac{{{2^{\frac{1}{2}}}}}{{{2^2}}}} = \log _{{2^3}}^{{2^{\frac{{ - 3}}{2}}}} = \frac{{ - \frac{3}{2}}}{3}\log _2^2 = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{ - \frac{1}{2}\log _2^2 =  - \frac{1}{2} \times 1 =  - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\log _{2\sqrt 2 }^{64} = \log _{2 \times {2^{\frac{1}{2}}}}^{{2^6}} = \log _{{2^{\frac{3}{2}}}}^{{2^6}} = \frac{6}{{\frac{3}{2}}}\log _2^2 = \frac{{12}}{3} \times 1 = 4}\end{array}} \right.$ $A =  - \frac{1}{2} \times 4 =  - 2$