SOLUTION: How would you solve: (a) 16^1/4 (b) 25^3/2 (c) 8^-2/3 ^ stands for power of Thanks.

Algebra ->  Square-cubic-other-roots -> SOLUTION: How would you solve: (a) 16^1/4 (b) 25^3/2 (c) 8^-2/3 ^ stands for power of Thanks.      Log On


   

Question 230724: How would you solve:
(a) 16^1/4 (b) 25^3/2 (c) 8^-2/3

^ stands for power of
Thanks.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How would you solve:
(a) 16^1/4 = 2
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(b) 25^3/2 = 5^3 = 125
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(c) 8^-2/3 = 2^-2 = (1/2)^2 = 1/4
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Cheers,
Stan H.