Variable Base Logarithm

$\log_{3x+5} {(9x^2+8x+8)} = \dfrac {\log {(9x^2+8x+8)}}{\log {(3x+5)} } > 2$

$\begin{aligned} \Rightarrow \log {(9x^2+8x+8)} & > 2 \log {(3x+5)} \\ 9x^2+8x+8 & > (3x+5)^2 \\ 9x^2+8x+8 & > 9x^2+30x+25 \\ 0 & > 22x+17 \\ x & < -\frac{17}{22} \end{aligned}$

We note that when $\log {(3x+5)} < 0$ then $\dfrac {\log {(9x^2+8x+8)}}{\log {(3x+5)} } < 0 \not{>} 2$ . Therefore, for $\dfrac {\log {(9x^2+8x+8)}}{\log {(3x+5)} } > 2$

$\Rightarrow \log{(3x+5)} > 0\quad \Rightarrow 3x+5 > 1 \quad \Rightarrow x > -\dfrac {4}{3}$

Therefore, the answer: $\boxed {x \in \left(-\frac{4}{3},-\frac{17}{22}\right)}$

The graph below clearly shows the answer.