<s>[INST] Logarithmic equation $\log_2(x+4)=\log_{4x+16}8$ So the problem goes:

What is the product of all solutions in the equation
  $$\log_2(x+4)=\log_{4x+16}8$$

The solution to this should be $31\over4$, but  I got $-14$. This is what I did:
\begin{align*}
\log_2(x+4) & = \log_{4x+16}8\\
\log_2(x+4) & = \frac{1}{\log_8(4x+16)}\\
\log_2(x+4) & = \frac{1}{\frac{1}{3}\log_2 4(x+4)}\\
\log_2(x+4) & = \frac{3}{\log_2 4+ \log_2(x+4)}\\
\log_2(x+4) & = \frac{3}{2 + \log_2(x+4)}\\
(\log_2(x+4))^2+2\log_2(x+4)-3 & =0\\
t^2+2t-3 & = 0 && \text{let $t = \log_2 (x + 4)$}
\end{align*} 
Then $t_1=3,\ t_2=-1$.  And from here, I get $x_1=4,\ x_2=-{7\over2}$
$x_1 \cdot x_2=-14$
I'm not sure where I messed up, or if there is a solution I didn't find.
 [/INST] It should be $t_1=1$ and $t_2=-3$ and the rest is true.
I got $x=-2$ or $x=-\frac{31}{8}.$
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