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?

$2^x + 4^x= 2$ $\Rightarrow$ $2^x (1 + 2^x ) = 2$  $\Rightarrow $ $1 + 2^x = 2 ^{1 - x}$ $\Rightarrow$ $1 + 2^x = 2 ^{- x} \times 2$
Now Set $ y = 2^x$; then we have
$1 + y = y^{-1} \times 2 $ $\Rightarrow$ $y^2 + y -2 = 0$. Which has solutions $y = 1$ and $y = -2$. 
$y = -2 $ is unacceptable, because $y = 2^x$ is a positive function. So $y = 2^x = 0$ is acceptable and will give us $ x = 0$. And we are done!
