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I am having trouble with this:

If $4^x + 4^{-x} = 34$, then what is $2^x + 2^{-x}$ equal to?

I managed to find $4^x$ and it is: $$4^x = 17 \pm 12\sqrt{2}$$ so that means that $2^x$ is: $$2^x = \pm \sqrt{17 \pm 12\sqrt{2}}.$$

Correct answer is 6 and I am not getting it :(. What am I doing wrong?

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I would like to say thank you to all of you who have helped me. You guys really helped me a lot :). Thank you for that. – Vizualni Jan 12 '11 at 22:05
up vote 9 down vote accepted

You haven't done anything wrong! To complete your answer, one way you can see the answer is $6$ is to guess that

$$17 + 12 \sqrt{2} = (a + b\sqrt{2})^2$$

Giving us

$$17 = a^2 + 2b^2, \ \ ab = 6$$

Giving us

$$a = 3, \ \ b = 2$$

Thus $$ \sqrt{17 + 12 \sqrt{2}} = 3 + 2\sqrt{2}$$

which gives $$2^x + 2^{-x} = 6$$

(And similarly for $17 - 12\sqrt{2}$)

A simpler way is to notice that $(2^x + 2^{-x})^2 = 4^x + 4^{-x} + 2$.

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Let $a=2^x$. Then $2^{-x}=1/a$ and $$\left(a+\frac1a\right)^2=a^2+2+1/a^2=2+(2^x)^2+(2^{-x})^2=2+4^x+4^{-x}=2+34=36.$$ This means that $a+1/a=6$ (it should be 6 or -6, since these are the square roots of $36$ but since $a$ and $1/a$ are positive, it is 6).

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$4^{x}+4^{-x} = 34$

$2^{2x}+2^{-2x} = 34$

Let $z= 2^x$.

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