Explanation:
remember :
(x^n)^m = x^(n×m)
and
1/x = x^‐1
therefore, also
1/(x^n) = (1/x)^n = x^-n
now, or situation is
xy = 1
y = 1/x = x^-1
we use this in the original equation
x^(1/a) = y^(1/5) = (1/x)^(1/5) = x^(‐1/5)
so, again
x^(1/a) = x^(‐1/5)
therefore
1/a = -1/5
5/a = -1
5 = -a
a = -5