Ответ:

1)

[tex]a = log_{2}( log_{ \sqrt{2} }16 ) [/tex]

[tex]a = log_{2}( log_{ {2}^{ \frac{1}{2} } } {2}^{4} ) [/tex]

[tex] \frac{4}{ \frac{1}{2} } log_{2}(2)[/tex]

[tex] \frac{4}{ \frac{1}{2} } \times 1[/tex]

[tex] \frac{4}{ \frac{1}{2} } = 8[/tex]

[tex]a = log_{2}(8) [/tex]

[tex]a = log_{2}( {2}^{3} ) [/tex]

[tex]3 log_{2}(2) [/tex]

[tex]3 \times 1[/tex]

[tex]3[/tex]

[tex]a = 3[/tex]

2)

[tex] log_{2 \times 3}(x - 4) + log_{2 \times 3}(x + 1) = 1[/tex]

[tex]x - 4 \leqslant 0 \\ x + 1 \leqslant 0[/tex]

[tex]x \leqslant 4 \\ x \leqslant - 1[/tex]

[tex]x\in < - \infty .4][/tex]

[tex] log_{2 \times 3}(x - 4) + log_{2 \times 3}(x + 1) = 1.x \in < 4. + \infty > [/tex]

[tex] log_{6}(x - 4) + log_{6}(x + 1) = 1[/tex]

[tex] log_{6}((x - 4)(x + 1)) = 1[/tex]

[tex] log_{6}(x x + x - 4x - 4) = 1[/tex]

[tex] log_{6}( {x}^{2} + x - 4x - 4 ) = 1[/tex]

[tex] {x}^{2} + x - 4x - 4 = {6}^{1} [/tex]

[tex] {x}^{2} + 1x - 4x - 4 = {6}^{1} [/tex]

[tex] {x}^{2} - 3x - 4 = 6[/tex]

[tex] {x}^{2} - 3x - 4 - 6 = 0[/tex]

[tex] {x}^{2} + 2x - 5x - 10 = 0[/tex]

[tex]x(x + 2) - 5(x + 2) = 0[/tex]

[tex](x + 2)(x - 5) = 0[/tex]

[tex]x + 2 = 0 \\ x - 5 = 0[/tex]

[tex]x + 2 - 2 = 0 - 2 \\ x - 5 + 5 = 0 + 5[/tex]

[tex]x = - 2 \\ x \in < 4. + \infty > \\ x = 5[/tex]

[tex]x = 5[/tex]