Radicals and Rational Exponents

gadawgs3

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Feb 15, 2013
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I need help solving this problem. I am trying to write d in terms of A and S.:confused:

S = 16Ad(-2/3)

This is what I have so far:

I divided both sides by 16A
Multiplied both sides by -3/2
Flipped the base in order to remove the negative exponent

16A3/2 = d
S3/2
 
I need help solving this problem. I am trying to write d in terms of A and S.:confused:

S = 16Ad(-2/3)

This is what I have so far:

I divided both sides by 16A
Good! Though it would be better to show what you got. Did you get \(\displaystyle \frac{S}{16A}= d^{-2/3}\)?

Multiplied both sides by -3/2
NOT Good! But I suspect you don't mean "multiplied" I suspect what you did was take the -3/2 power of each side.

Flipped the base in order to remove the negative exponent

16A3/2 = d
S3/2
"Flipped" is a very ambiguous word to use in a mathematics problem! "Inverted" would be better.

But your problem is in the second step. When you have \(\displaystyle \frac{S}{16A}= d^{-2/3}\) and took the -3/2 power, you did NOT do it to the "16".
\(\displaystyle d= \left(\frac{S}{16A}\right)^{-3/2}= \frac{S^{-3/2}}{16^{-3/2}A^{-3/2}}\). Yes, \(\displaystyle S^{-3/2}= \frac{1}{S^{3/2}}\) and \(\displaystyle \left(\frac{1}{A}\right)^{-3/2}= A^{3/2}\).

But what is \(\displaystyle \left(\frac{1}{16}\right)^{-3/2}\)?
 
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