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$\displaystyle \Rightarrow (1-i)^x=2^x$

$\displaystyle \Rightarrow (\sqrt{2} \cdot e^{\frac{-i\pi}{4}})^x = 2^x$

Taking modulus on both sides ,

$\displaystyle \Rightarrow |(\sqrt{2})^x| \cdot |(e^{\frac{-i\pi}{4}\cdot x}) | =|2^x|$

$\displaystyle \Rightarrow |(\sqrt{2})^x| \cdot (1) = |2^x|$

$\displaystyle \Rightarrow 1 = 2^{\frac{x}{2}}$

$\displaystyle \Rightarrow \boxed{x = 0}$

(1 - i)^x = 2^x implies 1- i = 2 which is untrue.

So original equation can only be true if each side = 1 which is true only if x = 0 since any n^0 = 1.

(1-i)^x=2^x => ((1-i)/2)^x=((1-i)/2)^0 Thus, x=0