Solve the equation: $1+2^x+4^x+8^x+16^x+32^x=3(1+2^x+4^x)$

Question

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...

Answer 1

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

Answer 2

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