[tex]log_3 (6-x) + 2log_3 (\sqrt[]{4-x}) =1[/tex], [tex]D = (- \infty; 4)[/tex]
[tex]2log_3 \sqrt{4-x} = log_3 ( \sqrt{4-x} )^2 = log_3 (4-x)[/tex]
[tex]log_3 (6-x) + log_3 (4-x) = log_3 [(6-x)(4-x)][/tex]
1 = [tex]log_3 3 [/tex]
[tex]log_3 [(6-x)(4-x)] = log_3 3 =\ \textgreater \ (6-x)(4-x) = 3 [/tex]
[tex]x^2 -10x +24 = 3 =\ \textgreater \ x^2 - 10x + 21 = 0 [/tex]
(x-3)(x-7) = 0 => x1 = 3 ∈ D ; x2 = 7 ∉ D
