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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)

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2 Answers

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Simply use the property of power :

$({x^a)}^b = x^{a*b}$

Particular case:

$({x^\alpha})^ {1/\alpha} = x$

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Simply use the property of power :

(x^a)^b = x^(a*b)

So in your case:

(x^alpha)^(1/alpha) = x
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