[tex]\it \begin{cases} \it a=\dfrac{log_248}{log_272}=\dfrac{log_2 2^4\cdot3}{log_2 2^3\cdot3^2}=\dfrac{4+log_2 3}{3+2log_2 3}\ \stackrel{not}{=}\ \dfrac{4+x}{3+2x}\\ \\ \\ \it b=\dfrac{log_2 24}{log_2 6}=\dfrac{log_2 2^3\cdot3}{log_2 2\cdot3}=\dfrac{3+log_2 3}{1+log_2 3}\ \stackrel{not}{=}\ \dfrac{3+x}{1+x} \end{cases}\ \ \ \ \ (*)[/tex]
[tex]\it a(b+3)-3b+1=0 \Leftrightarrow a(b+3)=3b-1 \stackrel{(*)}{\Leftrightarrow} \dfrac{4+x}{3+2x}\Big(\dfrac{3+x}{1+x}+3\Big)=\\ \\ \\ =\dfrac{9+3x}{1+x}-1 \Leftrightarrow\ \dfrac{4+x}{3+2x}\cdot\dfrac{6+4x}{1+x}=\dfrac{9+3x-1-x}{1+x}\Big|_{\cdot(1+x)}\ \Leftrightarrow\ \\ \\ \\ \ \Leftrightarrow\ \dfrac{(4+x)\cdot2(3+2x)}{3+2x}=8+2x\ \Leftrightarrow\ 8+2x=8+2x\ \ (A)[/tex]
