4
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ Hint: look at $8^x+16^x+32^x$ $\endgroup$ – Martigan Sep 12 '14 at 13:06
  • $\begingroup$ You lost a constant there at the end... $\endgroup$ – abiessu Sep 12 '14 at 13:08
11
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $0^0=1$, so $m=0$ is enough. $\endgroup$ – Thomas Andrews Sep 12 '14 at 13:12
  • $\begingroup$ @ThomasAndrews, en.wikipedia.org/wiki/Indeterminate_form#Indeterminate_form_00 $\endgroup$ – lab bhattacharjee Sep 12 '14 at 13:13
  • 1
    $\begingroup$ 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." $\endgroup$ – Thomas Andrews Sep 12 '14 at 13:15
  • $\begingroup$ 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! $\endgroup$ – Lukas Arvidsson Sep 12 '14 at 13:21
3
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.