Is there a way to simplify a^log n, a = 1 / b

[jstep]

Is there a way to simplify a^logn, a = 1 / b

I don't know if this is possible, but I thought I would ask for another opinion if I'm overlooking something. Anyway, is there a way to simplify a^log_bn, a = 1 / b

 

[MisterX]

yes

$\left(b^{-1}\right)^{log_{b} n} = \left(b^{log_{b} n}\right)^{-1} = n^{-1}$


as

$\left(b^{x}\right)^{y} = \left(b^{y}\right)^{x}$

 

[jstep]

aha! thanks!