A)
Firtly, we edd to simplify the plots.
[tex]3.9^x-5.6^x+2.4^x\ \textless \ 0 \\ \frac{3.3^{2x}-5.2^x.3^x+2.2^{2x}}{2^x.3^x} \ \textless \ 0 \\3.(\frac{3}{2})^x -5+2. (\frac{3}{2})^{-x}\ \textless \ 0 \\ 3.(\frac{3}{2})^x+ \frac{2}{(\frac{3}{2})^x} \ \textless \ 5[/tex]
[tex]3.(\frac{3}{2})^{2x}+ 2 \ \textless \ 5.(\frac{3}{2})^x \\ 3.((\frac{3}{2})^x)^2-5.(\frac{3}{2})^{x}+2\ \textless \ 0[/tex]
Making the substitution and equting to zero, we have:
[tex](\frac{3}{2})^x=y[/tex]
[tex] \\ \\ 3y^2-5y+2=0 \\ \\ \Delta=(-5)^2-4.3.2 \\ \Delta=25-25 \\ \Delta=1 \\ \\ [/tex]
[tex]x_{1}= \frac{5+1}{2.3} \\ x_{1}=1 \\ \\ x_{2}= \frac{5-1}{2.3} \\ x_{2}= \frac{2}{3} [/tex]
The result should be less than zero, soon:
[tex]\boxed {S=(xeR/1\ \textless \ x\ \textless \ \frac{2}{3} )}[/tex]
B)
Simply eliminating the logarithm, however, since the base is less than 1, we must reverse the inequality.
[tex]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)}[/tex]
If you notice any mistake in my English, please know me, because I am not native.
