Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$ I am doing some math repetition and am a bit stuck on this exercise:
Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$.
Now, this is a geometric sum on both the $LHS$ and $RHS$, which I guess is something that I should use to solve the equation... 
Another way is to simply start to eliminate terms:
$$1+2^x+4^x+8^x+16^x+32^x=3+3 \times 2^x+3\times4^x$$
$$-2 -2\times 2^x -2\times4^x+8^x+16^x+32^x = 0$$
$$8^x+16^x+32^x = 2 +2\times 2^x +2\times4^x$$
$$8^x+16^x+32^x = 2(1 + 2^x + 4^x)$$
But I am stuck here...

Setting $2^x=a,$
We have $$1+a+a^2+a^3(1+a+a^2)=3(1+a+a^2)$$
$$\iff (1+a+a^2)(a^3-2)=0$$
If $x$ is real $2^x>0\implies1+a+a^2>0$
So, we have $2=(2^x)^3\iff2^{3x-1}=1$
Now if $b^m=1$
either $m=0,b\ne0$
or $b=1$
or $b=-1,m$ is even
