Is anybody good at simplifying radical expressions
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Is anybody good at simplifying radical expressions

[From: ] [author: ] [Date: 11-09-16] [Hit: ]
=)-First you must rearrange the root, since it is impossible to compute 6th roots in your head. Recall Aroot(X^B) = X^B/A. 128 is a power of two, so figure out by what exponent you must multiply two by to get 128. By doing this the twos cancel and youre left with 1/x^1/6,......
the 6th root of 128x^7 divided by the 6th root of 2x? I have no idea how to do this

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= (128x^7/2x)^1/6 - 6th root of 128x^7 divided by the 6th root of 2x
= (64x^6)^1/6 - 6th root of 64x^6
=2
hope this will help you...

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alriii i m gonna tke on this question:

question: 6√(128^7 / 2x)

lets get the one of the 128 out of the 7 exponents to cancel out with the 6th root:
6√[(128^6) * (128) / (2x)]

then divide the 128 with the 2:
6√[(128^6) * (64) / (x)]

take out 64 to get similiar exponents:
6√[(128^6) * (2^6) /(x)]

therefore, the sixes cancel each other out:
(128)(2) / (6√x)
=256/(6√x)

=)

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First you must rearrange the root, since it is impossible to compute 6th roots in your head. Recall Aroot(X^B) = X^B/A. 128 is a power of two, so figure out by what exponent you must multiply two by to get 128. By doing this the twos cancel and you're left with 1/x^1/6, which = x^1/6. Set this equal to zero and use log laws to solve for x.
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