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?
Best Answer
The step where you go to $2 \cdot x + x + 1 = (\cdots)$ is incorrect, since $\log_4 (a+b) \neq \log_4 a + \log_4 b$ in general.