Problem (Algebra):
Determine all real values of $x$ such that

$$
\log _{5 x+9}\left(x^{2}+6 x+9\right)+\log _{x+3}\left(5 x^{2}+24 x+27\right)=4
$$

Solution:
We manipulate the given equation into a sequence of equivalent equations:

$$
\begin{array}{rll}
\log _{5 x+9}\left(x^{2}+6 x+9\right)+\log _{x+3}\left(5 x^{2}+24 x+27\right) & =4 & \\
\frac{\log \left(x^{2}+6 x+9\right)}{\log (5 x+9)}+\frac{\log \left(5 x^{2}+24 x+27\right)}{\log (x+3)} & =4 & \text { (using the "change of base" formula) } \\
\frac{\log \left((x+3)^{2}\right)}{\log (5 x+9)}+\frac{\log ((5 x+9)(x+3))}{\log (x+3)} & =4 & \text { (factoring) } \\
\frac{2 \log (x+3)}{\log (5 x+9)}+\frac{\log (5 x+9)+\log (x+3)}{\log (x+3)} & =4 & \text { (using logarithm rules) } \\
2\left(\frac{\log (x+3)}{\log (5 x+9)}\right)+\frac{\log (5 x+9)}{\log (x+3)}+\frac{\log (x+3)}{\log (x+3)} & =4 & \text { (rearranging fractions) }
\end{array}
$$



Making the substitution $t=\frac{\log (x+3)}{\log (5 x+9)}$, we obtain successively

$$
\begin{aligned}
2 t+\frac{1}{t}+1 & =4 \\
2 t^{2}+1+t & =4 t \\
2 t^{2}-3 t+1 & =0 \\
(2 t-1)(t-1) & =0
\end{aligned}
$$

Therefore, $t=1$ or $t=\frac{1}{2}$.

If $\frac{\log (x+3)}{\log (5 x+9)}=1$, then $\log (x+3)=\log (5 x+9)$ or $x+3=5 x+9$, which gives $4 x=-6$ or $x=-\frac{3}{2}$.

If $\frac{\log (x+3)}{\log (5 x+9)}=\frac{1}{2}$, then $2 \log (x+3)=\log (5 x+9)$ or $\log \left((x+3)^{2}\right)=\log (5 x+9)$ or $(x+3)^{2}=5 x+9$.

Here, $x^{2}+6 x+9=5 x+9$ or $x^{2}+x=0$ or $x(x+1)=0$, and so $x=0$ or $x=-1$.

Therefore, there are three possible values for $x: x=0, x=-1$ and $x=-\frac{3}{2}$.

We should check each of these in the original equation.

If $x=0$, the left side of the original equation is $\log _{9} 9+\log _{3} 27=1+3=4$.

If $x=-1$, the left side of the original equation is $\log _{4} 4+\log _{2} 8=1+3=4$.

If $x=-\frac{3}{2}$, the left side of the original equation is $\log _{3 / 2}(9 / 4)+\log _{3 / 2}(9 / 4)=2+2=4$.

Therefore, the solutions are $x=0,-1,-\frac{3}{2}$.

Final answer: $0,-1,-\frac{3}{2}$