Maths question help..

Ohnoezfail

Ohnoezfail

Ohnoezfail

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Joined
30 Nov 2008
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253
Okay.. I have my higher gcse retake exams tommorow, and I need OcUK's help;)
The question is to write 27^-1/3 as a fraction
But I don't think I'm getting the correct answer :o
Here are my steps:

27^-1/3


1
-----------
sqrt27 ^1/3

1
------------
3sqrt27 ^1/3


3sqrt27 = 9sqrt3


1
------------
9sqrt3 ^1/3


9sqrt3 * 1/3 = 3sqrt3


Any geniuses help me out here please?
 
27^(1/3) would be cube root of 27, 3 x 3 x 3 = 27 so the cube root of 27 is 3.

Because in your question its 27^(-1/3) it's equal to 1/27^(1/3) so its 1/3.
 
As pointed out the solution is 1/3

The rule is with a fraction m/n you raise the number to the mth power and then take the nth root (or the other way round if it suits you, it may be easier as the numbers will be smaller).

So to take another example 32^(2/5) = 4. The 5th root of 32 is 2 and squaring gives 4.

When adding a minus sign you just take the reciprocal of the number, that is put 1/n. For example 16^(-3/2) is 1/64.
 
dante6491 said:
To the power of M/N you always do the N part first, it wouldn't work otherwise.
Click to expand...

What? It's a commutative operation, as you can think of M/N as the product (M)(1/N). Since multiplication is a commutative operation (a^m)^(1/n) = (a^(1/n))^m, it doesn't matter whether you do the M part or the N part first.

For example in the previous one, 32^(2/5):

(32^2)^(1/5) = (1024)^(1/5) = 4
(32^(1/5))^2 = (2)^2 = 4

It's usually just simpler to do the 1/N part first, if you're doing the calculation in your head.
 
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