Compute modular inverse

Python, 29 bytes

lambda x,n:pow(x,2**n-1,2**n)

This returns 0 for even x. It uses Euler’s theorem, with the observation that 2^n − 1 is divisible by 2^(n − 1) − 1, via Python’s builtin fast modular exponentiation. This is plenty fast enough for n up to 7000 or so, where it starts taking more than about a second.

Python 3.8 (pre-release), 25 bytes

lambda a,b:pow(a,-1,2**b)

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Mathematica - 22

f=PowerMod[#,-1,2^#2]&

f[x,n] returns y with x*y=1 mod 2^n, otherwise x is not invertible modulo 2^n