I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$
I've tried this far, and I'm stuck
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x (x+12)- \log_x 4^3 &= -1 \\
\log_x(x+12)- \log_x 64 &= -1 \end{align}$$
then I substituted $4^{y+3x} = 64$
$\log_x (x+12) - \log_x 4^{y+3x} = -1$
I don't know what should I do next. any ideas?

$4^{y+3x} = 64$
$\log_4 ({4^{y+3x}}) = \log_4 (64)$
$(y+3x)\log_4 {4} = 3$
$y + 3x = 3$

$\log_x (x+12) - 3\log_x (4) = -1$
$\log_x (\frac{x+12}{4^3}) = -1$
$\frac{x+12}{64} = x^{-1}$
$x^2 + 12x - 64 = 0$
$(x+16)(x-4) = 0$
$x = 4$, as $x > 1$ for $\log_x (x+12) - 3\log_x (4) = -1$
(Why must the base of a logarithm be a positive real number not equal to 1?)

$y + 3(4) = 3$
$y = -9$

$4 + 2(-9) = -14$
