${(32)^{{{\tan }^2}x}} + {(32)^{{{\sec }^2}x}} = 81$ $\Rightarrow {(32)^{{{\tan }^2}x}} + {(32)^{1 + {{\tan }^2}x}} = 81$ $\Rightarrow {(32)^{{{\tan }^2}x}} = {{81} \over {33}}$ taking log of base 32 both side, $\Rightarrow$ tan 2 x = ${\log _{32}}\left( {{{81} \over {33}}} \right)$ $\Rightarrow$ tan x = $\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)}$ As value of $\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)}$ belongs to (0, 1). In interval $\,0 \le x \le {\pi \over 4}$ only one solution.