\( log_{_{0.3}}(2x+3) \leq log_{_{0.3}}(x-1) \)

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.

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