Posted by **Eve** on Thursday, September 10, 2009 at 8:18pm.

I am trying to figure out how to evaluate 100^-3/2. I'm going by the example (64)^2/3=16. IN the example, 64^2/3 is changed into 64^1/3^2=4^2=16. But I don't get why out of all numbers it is changed into 4^2. I know 16 goes into 64 four times, but why is the answer 16 and what becomes of the 1/3? So how would this apply to 11^-3/2? Thanks for your help!

- Algebra -
**MathMate**, Thursday, September 10, 2009 at 8:38pm
First, you'd have to adhere to the rules of priority.

The expression 64^{2/3} is much clearer.

The 2 in the exponent requires us to square the base, and the 3 in the denominator represents (1/3) means we need to take the cube root of the base.

So whether you do ∛((64)²) or (∛64)³ will give the same answer, i.e. ∛(4096)=16 or (4)²=16. The 4 comes from the cube root of 64, or 64^{1/3}.

For 11^{-3/2} it is the same procedure, by following the laws of exponents:

x^{a+b} = x^{a} × x^{b}

x^{ab} = (x^{a})^{b}

x^{-a} = 1/x^{a}

x^{1/a} = ath root of x.

Therefore

11^{-3/2}

=1/11^{3/2}

=1/∛(11³)

=1/∛(1331)

- Algebra-correction -
**MathMate**, Thursday, September 10, 2009 at 8:39pm
11^{-3/2}

=1/11^{3/2}

=1/√(11³)

=1/√(1331)

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