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.


\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?

  • 5
    $\begingroup$ 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. $\endgroup$
    – Martin R
    Feb 24, 2022 at 9:21
  • 2
    $\begingroup$ Now the problem has been identified, you can write the original problem as a quadratic in $y:=4^x$ to solve it. $\endgroup$
    – J.G.
    Feb 24, 2022 at 9:27
  • 2
    $\begingroup$ 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. $\endgroup$
    – tkf
    Feb 24, 2022 at 22:49

3 Answers 3


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.


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 $$x=\frac{1}{2}.$$


There is a mistake going from the 4th to the 5th line: $4^a+4^b\neq4^{a+b}$


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