$\log _2^{4{x^2}} = \log _2^{{2^2}{x^2}} = \log _2^{\left( {2x} \right)} = 2\log _2^{2x}$$2\log _2^{2x} + \frac{6}{{\log _2^{2x}}} = 8 \Rightarrow $ مخرج مشترک می‌گیریم$2{\left( {\log _2^{2x}} \right)^2} + 6 = 8\log _2^{2x}$$ \Rightarrow 2{\left( {\log _2^{2x}} \right)^2} - 8\log _2^{2x} + 6 = 0$$ \Rightarrow \log _2^{2x} = A \Rightarrow 2{A^2} - 8A + 6$$ = 0 \to  \div 2 = {A^2} - 4A + 3 = 0$$ \Rightarrow \left( {A - 1} \right)\left( {A - 3} \right) = 0 \Rightarrow A - 1 = 0 \Rightarrow A = 1$$ \Rightarrow \log _2^{2x} = 1 \Rightarrow 2x = 2 \Rightarrow x = 1$$A - 3 = 0 \Rightarrow A = 3 \Rightarrow \log _2^{2x} = 3$$ \Rightarrow 2x = {2^3} \Rightarrow 2x = 8 \Rightarrow x = 4$جمع ریشه‌ها ${x_1} + {x_2} = 1 + 4 = 5$