Expert: Ahmed Salami - 11/11/2004

Question
i have trouble dealing with the negatives on the exponents.

1) (3^-2x)^2 * x^-5
2) x^-1(x^-1/3 - x^2/3) / x^-4/3

Answer
Hi Amy,
Sorry about the delay, a technical problem.
First let me state that
(x^a)^b = x^(ab)
(x^a)(x^b) = x^(a+b)
(x^a)/(x^b)= x^(a-b)
x^-a can be regarded as 1/(x^a)
1)(3^-2x)^2 = 3^-4x
(3^-2x)^2 * x^-5 = (3^-4x)(x^-5)
or 1/[(3^4x)(x^5)]
You van also convert 3^4x to (3^4)^x = 81^x
Therefore,
(3^-2x)^2 * x^-5 = 1/[(81^x)(x^5)]

2)(x^-1)(x^-1/3 - x^2/3) / (x^-4/3)  
= (x^-1)(x^-1/3) - (x^-1)(x^2/3) / x^-4/3  
= [x^(-1 + -1/3) - x^(-1 + 2/3)]/ x^-4/3
= [x^(-4/3) - x^(-1/3)]/ x^-4/3
= [x^(-4/3) / x^(-4/3)]  -  [x^(-1/3) / x^(-4/3)]
= 1 - x^[(-1/3) - (-4/3)]
= 1 - x^(-1/3 + 4/3)
= 1 - x^1
= 1 - x

I hope you understand it. Once again, i'm sorry for the delay.
Regards.