<s>[INST] Solution for $\log_{3}(2x+5) = [\log_{9}(4x+1)]^2$ This was my logic:
$$
\log_{3}(2x+5) = [\log_{9}(4x+1)]^2 \iff 
 \log_{3}(2x+5) = \left(\dfrac{\log_{3}(4x+1)}{\log_{3}9}\right)^2 \iff
 \log_{3}(2x+5) = \dfrac{\log_{3}(4x+1) \cdot \log_{3}(4x+1)}{4} \iff
 4 \cdot \log_{3}(2x+5) - \log_{3}(4x+1) \cdot \log_{3}(4x+1) = 0
$$
But now factorization isn't helping me. I don't see if there is a way to join the logs or if there is a convenient substitution. Probably there is a typo at the source of the equation, but I got curious and wanted to solve this.
ps: the entire log is squared, and it turns hard to solve for that reason.
 [/INST] Alternatively:
$$\log_{3} (2x+5)=\log_{3^2} (4x+1)^2=\frac{2}{2} \log_{3} |4x+1| \Rightarrow \\
\pm(2x+5)=4x+1 \Rightarrow \\
x=2;-1.$$
</s>