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What is the cubed root of W?

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by DavidG@VeritasPrep » Sun Mar 01, 2015 6:08 pm
First, the xth root of w will just be w^(1/x)


1) the 5th root of W is 64.
This tells us that w^(1/5) = 64. So w = 64^5. We have w, so sufficient.



2) the 15th root of W is 4
This tells us that w^(1/15) = 4. So w = 4^15. Again, we have w, so sufficient.
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by DavidG@VeritasPrep » Sun Mar 01, 2015 6:41 pm
And if we actually wanted to solve for the cubed root of w, or w^(1/3):

S1: w = 64^5. Because 64 = 2^6, we can rewrite 64^5 as (2^6)^5 = 2^30.
(2^30)^(1/3) = 2^10


S2: w = 4^15, we can rewrite as (2^2)^15 = 2^30.
Again, (2^30)^(1/3) = 2^10

But for DS, it's enough to see that if we can solve for w, we can get a unique value for w^(1/3)
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by Matt@VeritasPrep » Mon Mar 02, 2015 3:57 pm
As for how to solve each one, the algebra looks like this:

S1:: �√w = 64. Raising both sides to the fifth gives w = 64�.
S2:: ¹�√w = 4. Raising both sides to the fifteenth gives w = 4¹�.

Notice too that these are the same number: 64� = (4³)� = 4¹�.
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