SOLUTION: I need help on and with these problems if you could help that would be wonderful 8^0 * 2^-3 (2^3)^-2 (6^-1)^2 x^-8 3x^-5

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: I need help on and with these problems if you could help that would be wonderful 8^0 * 2^-3 (2^3)^-2 (6^-1)^2 x^-8 3x^-5      Log On


   

Question 1152502: I need help on and with these problems if you could help that would be wonderful
8^0 * 2^-3
(2^3)^-2
(6^-1)^2
x^-8
3x^-5
Found 3 solutions by josgarithmetic, MathLover1, Theo:
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
only doing the first one

8^0 * 2^-3

8%5E0+%2A+2%5E-3
1%2A%281%2F2%5E3%29
1%2F8

Answer by MathLover1(20850) About Me  (Show Source):
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
8^0 * 2^-3 = 1 * 1/2^3 = 1 * 1/8 = 1/8
(2^3)^-2 = (1/(2^3)^2 = 1/2^6 = 1/64
(6^-1)^2= ) = (1/6)^2 = 1/36
x^-8 = 1/x^8
3x^-58^0 * 2^-3 ***** see below the rest of these.
(2^3)^-2 = 8^-2 = 1/8^2 = 1/64
(6^-1)^2 = 1/6^1)^2 = (1/6)^2 = 1/36
x^-8 = 1/x^8
3x^-5 = 3/x^5

3x^-58^0 * 2^-3 = 3x^(-58^0) * 2^-3 = 3x^(-1) * 1/2^3 = 3/x^1 * 1/8 = 3/8x

i checked these answers out and they appear to be good.

some concepts.

here's a reference on exponent arithmetic that you might find helpful.

https://www.purplemath.com/modules/exponent.htm