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.
 A: I might be wrong, but $$\log_3 (2x+5)=\log_9 (4x+1)^2 = 2\log_9 (4x+1)$$ implies $$2x+5=3^{2\log_9 (4x+1)}=(3^2)^{\log_9 (4x+1)}=9^{\log_9 (4x+1)}=4x+1.$$ 
A: Note that $\log_a (b)^2=2\log_a (b)=2\cdot\dfrac{\log b}{\log a}\ne\left(\dfrac{\log b}{\log a}\right)^2$
Apply this to your second step, which I believe you performed incorrectly.
If you do this correctly, you will eventually reach that:

$\log_3(2x+5)=2\cdot\dfrac{\log_3(4x+1)}{\log_3 9}$

A: Alternatively:
$$\log_{3} (2x+5)=\log_{3^2} (4x+1)^2=\frac{2}{2} \log_{3} |4x+1| \Rightarrow \\
\pm(2x+5)=4x+1 \Rightarrow \\
x=2;-1.$$
A: After search a while, it seems like only numerical methods can provide the answer to my question. Although I was wanting some symbolic manipulation, that seems the way for now.
A: Since you loog for the solution of $$\log_{3}(2x+5) = \left(\log_{9}(4x+1)\right)^2$$ (let me switch to natural logarithms) consider that you look for the zero of function
$$f(x)=\frac{\log (2 x+5)}{\log (3)}-\frac{\log ^2(4 x+1)}{\log ^2(9)}$$
$$f'(x)=\frac{2}{(2 x+5) \log (3)}-\frac{8 \log (4 x+1)}{(4 x+1) \log ^2(9)}$$ Use inspection and after a "few" attempts, you will see that $$f(10)=\frac{\log (25)}{\log (3)}-\frac{\log ^2(41)}{\log ^2(9)}\approx +0.0734432$$ $$f(11)=\frac{\log (27)}{\log (3)}-\frac{\log ^2(45)}{\log ^2(9)}\approx -0.0015104$$
So, the solution is very close to $x_0=11$. Since there is no analytical solution, use Newton method and get the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 11.00000000 \\
 1 & 10.97923902 \\
 2 & 10.97925143
\end{array}
\right)$$ If you perform only one ieration of Newton method, you would get
$$x_1=11-\frac{135} 2\frac{ \log ^2(9) \log (27)-\log (3) \log ^2(45)}{ 5 \log
   ^2(9)-12 \log (3) \log (45)}$$ that is to say
$$x_1=11-\frac{135} 8 \frac{8 \log ^2(3)-\log ^2(5)-4 \log (3) \log (5)}{\log (375)}$$
