<s>[INST] 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...
 [/INST] Note that the left side of the equation can be written as
$2^0+2^x+2^{2x}+2^{3x}...2^{5x}$
This is a geometric series, with 
a=1, n=6, and $r=2^x$
We use the formula:
$S_n = \frac {a(1-r^n)}{1-r}$
Substitute the values, you get 
$S=\frac{(1-2^{6x})}{1-2^x}$
We do the same for the right side
$S=3[\frac{1-2^{3x}}{1-2^x}]$
Equate the terms, and rearrange
$\frac{(1-2^{6x})}{1-2^x}=3[\frac{1-2^{3x}}{1-2^x}]$
$1-2^{6x}=3-3\cdot2^{3x}$
$-2^{6x}+3\cdot2^{3x}-2=0$
And this bit is my fave.
It's just a quadratic!
Because $-2^{6x} = -(2^{3x})^2$
Now you just let $2^{3x}$ = u
$-u^2+3u-2=0$
Solve for u
Then you just sub back $2^{3x}$
And there you have it.
If I have made an error (as I am prone to), notify me and I will withdraw my answer.
Thanks
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