Привет) как это решить? :D

1)
$\sqrt{ 2^{x} } * \sqrt{ 3^{x} } = \sqrt{ 6^{x} } = 36 \\ 6^{x}=36^{2} \\ 6^{x}=6^{4} \\ x=4$

2)
$4^{x}=64 \\ 4^{x}=4^3 \\ x=3$

3)
$5^{x^{2}-2x-1}=5^{2} \\ x^{2}-2x-1=2 \\ x^{2} -2x -3 =0 \\ x=3, x=-1$

4)
64 \\ 2^{5x-8}> 2^{6} \\ 5x-8>6 \\ 5x>14 \\ x>2.8" alt="2^{5x-8}>64 \\ 2^{5x-8}> 2^{6} \\ 5x-8>6 \\ 5x>14 \\ x>2.8" align="absmiddle" class="latex-formula">

5)
$8^{x+1} = 0.25 \\ 8^{x}*8= \frac{2}{8}\\ 8^{x}= \frac{2}{8} \times \frac{1}{8} \\ 2^{3x}=\frac{1}{32} \\ 2^{3x}=2^{-5} \\ 3x=-5 \\ x=-\frac{5}{3} \\ x=-1\frac{2}{3}$

6)
$3^{3x-3}=27 \\ 3^{3x-3}=3^3 \\ 3x-3=3 \\ 3x=6 \\ x=2$

7)
$5^{x^{2}-6x} = 5^{-3x+8} \\ x^{2}-3x-8=0 \\ D<0$

8)
$log_{ \sqrt{2} }(x+1)=2 \\ (\sqrt{2})^{2} = x+1 \\ x+1=2 \\ x=1$

9)
$lg(x(x-3)-lg\frac{x-3}{4x} = 0 \\ lg\frac{x(x-3)4x}{x-3}=0 \\ 4x^{2}=1 \\ x^2=\frac{1}{4} \\ x=\frac{1}{2}$