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?

Consider $$\log_x(x+12)- 3 \log_x(4)= -1$$ and change base to get $$\frac{\log (x+12)}{\log (x)}-\frac{3 \log (4)}{\log (x)}=-1$$ Assuming $x\neq 1$, multiply each term by $\log (x)$ to get $$\log (x+12)-3 \log (4)=-\log (x)$$ thatt is to say $$\log (x+12)+\log(x)=\log(4^3)=\log(64)$$ $$\log\left(x(x+12)\right)=\log(64)$$ $$x(x+12)=64\implies x^2+12x-64=0$$ the roots of which geing $-16$ (to be discarded since $x$ must be positive) and $x=4$ which is then the only root.
Then $y$ from what you already established.
