Which principle is violated here in the logarithmic equation?

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:

$$$$16^{x}+4^{(x+1)}=12$$$$

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}

• If you get a wrong solution then you can simply go back and substitute your solution in the previous equations. Then you'll quickly find the point where you made an error. Commented Feb 24, 2022 at 9:21
• Now the problem has been identified, you can write the original problem as a quadratic in $y:=4^x$ to solve it.
– J.G.
Commented Feb 24, 2022 at 9:27
• Mistakenly claiming that $f(a+b)=f(a)+f(b)$ for some function $f$, is one of the most common errors in basic mathematics. In your case $f(x)=\log(x)$. Other common examples are $f(x)=\frac 1x$ and most famously $f(x)=x^n$ - see: en.wikipedia.org/wiki/Freshman%27s_dream.
– tkf
Commented Feb 24, 2022 at 22:49

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.
\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 $$x=\frac{1}{2}.$$
There is a mistake going from the 4th to the 5th line: $$4^a+4^b\neq4^{a+b}$$