# 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...

• Hint: look at $8^x+16^x+32^x$ – Martigan Sep 12 '14 at 13:06
• You lost a constant there at the end... – abiessu Sep 12 '14 at 13:08

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

• $0^0=1$, so $m=0$ is enough. – Thomas Andrews Sep 12 '14 at 13:12
• – lab bhattacharjee Sep 12 '14 at 13:13
• Indeterminate does not mean undefined. It only means that the limit $$\lim_{(x,y)\to (0,0)} x^y$$ doesn't exist. There is a reason we use the word "indeterminate" rather than "undefined." – Thomas Andrews Sep 12 '14 at 13:15
• I am not sure that I understand this completely. Running the equation on Wolfram gives me the (real) root $\frac{1}{3}$. By using the substitution method $2^x = a$ we get: $$a^3+a^4+a^5 = 2(1 + a + a^2) \implies a^3(1+a+a^2) = 2(1 + a + a^2)$$ This gives me that $a^3 = 2$, which is correct now that I think about it. Thank you! – Lukas Arvidsson Sep 12 '14 at 13:21

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.

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

• Thank you for your answer! Much apreciated! – Lukas Arvidsson Sep 12 '14 at 13:48
• It works? they said I was crazy!! You're very welcome – surelyourejoking Sep 12 '14 at 13:49
• I am not sure, solving the quadratic function gives me two roots: $u = 1$ and $u = 2$. Only $u = 2$ is right. Any ideas? – Lukas Arvidsson Sep 12 '14 at 13:58