Why does 2^(-log₂(x)) simplify to 1/x?

[repugno]

2^(-log2(x)) It reads, 2 to the power of -log base2 x

The problem is that I don't understand why this can also be written as x^-1

For some reason the base and the log2 cancel out. Can anyone explain to me why this happens, please?

1/2^(log2(x)) = 1/x

 

[arildno]

-log2(x)=log2(x^(-1)),
by the rule for logarithms:
blog(a)=log(a^(b))

 

[Zurtex]

Or you could use that:

$$a^{bc} = \left(a^c \right)^b$$

As $- \log_2 x = (-1) \log_2 x$