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Figuring out the fundamental

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Figuring out the fundamental

by rkav » Thu Mar 14, 2013 4:53 pm
Problem 15 pg 276 OG 13th edition.

In this Ds problem I understand why the answer is "D" Each statement alone is sufficient, but had a question on how they broke out the fundamentals in the answer key-

What is the cube root of 'w'?
1. the 5th root of w is 64
2. the 15th root of w is 4


Answer explanation- given the 5th root of w is 64 then w^(1/5) is 64 and (w^1/5)^5/3 + 64 ^ 5/3

I am not understanding why they raised the (w^1/5) to the (5/3) where does this 5/3 come from?


Thanks!
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by srcc25anu » Thu Mar 14, 2013 5:45 pm
Cube root of W means W^(1/3)
question asks what is W^(1/3)???
1. 5th root of w = 64 i.e. W^(1/5) = 64
if we raise both sides to the power of 5 then w = 64^5
sinec we need to find out W^(1/3) now we take the cube root on both sides that gived us W^(1/3) = {(64^5)^(1/3)} or 64^(5/3)
64 can be written as 2^6 so 64^(5/3) = 2^(6^5/3) or 2^10 hence sufficient

2. 15th root of W or w^(1/15) = 4
Raising both sides by 15th power: w = 4^15
Now taking the 3rd root on eavh side: w^(1/3) = 4^5 or 2^10 hence sufficient

Answer D
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by rkav » Thu Mar 14, 2013 5:58 pm
thank you! makes sense. I did not set it up as an equality which is where problem occured.
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