TITLE: Which principle is violated here in the logarithmic equation?
QUESTION [5 upvotes]: I'm a beginner to mathematics and I'm stuck with a calculus exercise. It seems like I violate a principle, but I cannot yet see what I did wrong here. I hope a second look from a 3rd person will help.
The equation I need to solve is:
\begin{equation}
16^{x}+4^{(x+1)}=12
\end{equation}
With as $x=\frac{1}{2}$ as only real solution. I understand that this can be solved with substitution, but I especially want to solve this problem with writing everything in the same base.
Attempt:
\begin{aligned}
&16^{x}+4^{(x+1)}=12 \\
&\left(4^{2}\right)^{x}+4^{(x+1)}=12 \\
&4^{2 \cdot x}+4^{(x+1)}=12 \\
&4^{2 \cdot x}+4^{(x+1)}=4^{\left(\frac{\log (12)}{\log (4)}\right)} \\
&2 \cdot x+x+1=\frac{\log (12)}{\log (4)} \\
&3 \cdot x=\frac{\log (12)}{\log (4)}-1 \\
&x=\frac{1}{3}\left(\frac{\log (12)}{\log (4)}-1\right)
\end{aligned}
Could I please get feedback?

REPLY [12 votes]: Other answers showed the error in your  method. Here is a correct method;
$$
\begin{aligned}
&16^{x}+4^{(x+1)}=12 \\
&\left(4^{2}\right)^{x}+4^{(x+1)}=12 \\
&\left(4^{x}\right)^{2}+4\cdot 4^{x}-12=0 \\&y^2 + 4y - 12 = 0,\\
\end{aligned}$$
which is now a quadratic equation with $y:=4^x$. Solving gives either $y=2$ or $y=-6$, so we now have $$4^x = 2 \qquad\mathrm{or}\qquad 4^x=-6.$$ Given that you are looking only for real solutions, we conclude that $4^x=2$ and $$x=\frac{1}{2}.$$