2) $$3\cdot 9^x-5\cdot 6^x+2\cdot 4^x < 0$$
$$log_{_{0.3}}(2x+3) \leq log_{_{0.3}}(x-1)$$

A) Firtly, we edd to simplify the plots.
$$3.9^x-5.6^x+2.4^x\ < \ 0 \\ \frac{3.3^{2x}-5.2^x. 3^x+2.2^{2x}}{2^x. 3^x} \ < \ 0 \\3.(\frac{3}{2})^x -5+2. (\frac{3}{2})^{-x}\ < \ 0 \\ 3.(\frac{3}{2})^x+ \frac{2}{(\frac{3}{2})^x} \ < \ 5$$
$$3.(\frac{3}{2})^{2x}+ 2 \ < \ 5.(\frac{3}{2})^x \\ 3.((\frac{3}{2})^x)^2-5.(\frac{3}{2})^{x}+2\ < \ 0$$

Making the substitution and equting to zero, we have:

$$(\frac{3}{2})^x=y$$
$$\\ \\ 3y^2-5y+2=0 \\ \\ \Delta=(-5)^2-4.3.2 \\ \Delta=25-25 \\ \Delta=1 \\ \\$$
$$x_{1}= \frac{5+1}{2.3} \\ x_{1}=1 \\ \\ x_{2}= \frac{5-1}{2.3} \\ x_{2}= \frac{2}{3}$$

The result should be less than zero, soon:

$$\boxed {S=(xeR/1\ < \ x\ < \ \frac{2}{3} )}$$
B)

Simply eliminating the logarithm, however, since the base is less than 1, we must reverse the inequality.

$$log_{_{0.3}}(2x+3) \leq log_{_{0.3}}(x-1) \\ 2x+3 \geq x-1 \\ x \geq -4 \\ \\ \boxed {S=(xeR|x \geq -4)}$$

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