$${\left( {32 \times {{10}^{ - 5}}} \right)^{ 2}} \times $$    $$64\, \div $$ $$\left( {{2^{16}} \times {{10}^{ - 4}}} \right)$$   $$ = $$ $${10^?}$$

Solution (By Examveda Team)

$$\eqalign{ & {\left( {32 \times {{10}^{ - 5}}} \right)^{ 2}} \times 64 \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) = {10^?} \cr & \Rightarrow {\left( {{2^5} \times {{10}^{ - 5}}} \right)^{ 2}} \times {2^6} \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) \cr & \,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right] \cr & \Rightarrow \frac{{{2^{10}} \times {{10}^{ - 10}} \times {2^6}}}{{{2^{16}} \times {{10}^{ - 4}}}} \cr & \,\,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow \frac{{{2^{16}} \times {{10}^4}}}{{{2^{16}} \times {{10}^{10}}}} \cr & \,\,\,\,\,\,\,\, = {10^?}.....\left[ {{a^{ - m}} = \frac{1}{{{a^m}}}} \right] \cr & \Rightarrow {10^{4 - 10}} = {10^?} \cr & \Rightarrow {10^{ - 6}} = {10^?} \cr & \Rightarrow ? = - 6 \cr} $$