I would say, take the root, but here the power is $-\alpha$.
In my book about the level of optimal employment (L), they go from
$\tag{1}((1-\alpha)/(1+tp))*A*((K/L)^{\alpha}) = W/P$
to
$\tag{2} L = (((W/P(1+tp))/((1-\alpha)*A))^{-1/\alpha})*K$
So they solved for L.
Sorry, I don't know how to make nice equations in this forum (corrected)
$\endgroup$ 12 Answers
$\begingroup$Simply use the property of power :
$({x^a)}^b = x^{a*b}$
Particular case:
$({x^\alpha})^ {1/\alpha} = x$
$\endgroup$ $\begingroup$Simply use the property of power :
(x^a)^b = x^(a*b)So in your case:
(x^alpha)^(1/alpha) = x $\endgroup$