Подскажите как правильно решить )log²(4-3x)-log²16=3 2)log³3x+log³6<3 3)log9(x²-2x-15)>1

1)
$log_{2}(4 - 3x) - log_{2}(16) = 3 \\ log_{2}( \frac{4 - 3x}{16} ) = log_{2}8 \\ \frac{4 - 3x}{16} = 8 \\ x = 41.3$
2)
$log_{3}3x + log_{3}6x < 3 \\ log_{3}(3x \times 6x) < log_{3}27 \\ 3x \times 6x < 27 \\ 18 {x}^{2} < 27 \\ x ^{2} < \frac{3}{2} \\ x < \sqrt{ \frac{3}{2} } \\$
3)
1 \\ log_{?}(x ^{2} - 2x - 15) > log_{9}9 \\ {x}^{2} - 2x - 15 > 9 \\ (x + 4)(x - 6) > 0 \\ " alt=" log_{9}( {x}^{2} - 2x - 15) > 1 \\ log_{?}(x ^{2} - 2x - 15) > log_{9}9 \\ {x}^{2} - 2x - 15 > 9 \\ (x + 4)(x - 6) > 0 \\ " align="absmiddle" class="latex-formula">
ответ (-беск;-4)U(6;+беск)