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$2^{2048x}+2^{-2048x}=2\\ 2^{2048x}+\dfrac{1}{2^{2048x}}=2$

Let $a=2^{2048x}$ . Substitute this in:

$a+\dfrac{1}{a}=2\\ a^2+1=2a\\ a^2-2a+1=0\\ (a-1)^2=0\\ a=1\\ 2^{2048x}=1=2^0\\ 2048x= 0\\ x=0$

Therefore, $2^x+2^{-x}=2^0+2^{-0}=1+1=\boxed{2}$

Using the AM-GM inequality, $\large \dfrac{ 2^{2048x}+\dfrac{1}{2^{2048x}} }{2} \le \sqrt{ \dfrac{2^{2048x}}{2^{2048x}} } \, \Rightarrow \, 2^{2048x}+\dfrac{1}{2^{2048x}} \le 2$ Equality occurs when $2^{2048x} = \dfrac{1}{2^{2048x}}$ , so $\large 2^{2048x}+\dfrac{1}{2^{2048x}}=2 \, \Rightarrow \, 2^{2048x} = \dfrac{1}{2^{2048x}} \, \Rightarrow \, 2^{4096x}=1 \, \Rightarrow \, x=0$ Thus $2^x+2^{-x}=2^0+2^{-0}=\large \color{#20A900}{\boxed{2}}$ .

$2^{2048x} + 2^{-2048x} = 2$

$2^{2048x} + 2^{2048x \times - 1} = 2$

$2^{2048x} + \frac {1}{ 2^{2048x}} = 2$

We can see that we need to equate a large exponent to that of $2^1$ .

In order to do so, the variable $x$ must be very small such that the left hand side will equate to the right.

The only value of $x$ is 0 as $2^1 = 2^0 + 2^0$

Hence the value of $2^x +2^{-x} = 2$