Solve $2^x+4^x=2$ This is the equation, but the result is different from wolframalpha:
$$2^x+4^x=2$$
$$2^x+2^{2x}=2^1$$
$$x+2x=1$$
$$x=\frac{1}{3}$$
WolframAlpha: $x=0$
Where is the error?

$$2^x+2^{2x}=2$$
Now put $2^x=t$
$$t+t^{2}=2$$
$$t^{2}+t-2=0$$
$$(t-1)(t+2)=0$$
Thus $t=1$ or $t=-2$
$2^x=1$ or $2^x=-2$
Since $2^x>0 $ for all real $x$ , $2^x=1=2^0$
Therefore $x=0$
