SOLUTION: Can you please help with the following : Simplify the expression. Use positive exponent only: (4𝑥^−1)^2(2𝑥^3)^3 Thanks

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Question 1093853: Can you please help with the following :
Simplify the expression. Use positive exponent only: (4𝑥^−1)^2(2𝑥^3)^3
Thanks

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i see this problem as:

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

this is equal to 4^2 * x^(-2) * 2^3 * x^9

rearrange the terms to get 4^2 * 2^3 * x^(-2) * x^9

simplify further to get 128 * x^(-2) * x^9

combine like terms to get 128 * x^7

that's your solution.

you can confirm using your calculator by evaluating the original expression and the final expression using a random value of x.

if the answer comes out the same, then you did good.

i used x = 25 and i got both expressions equal to 7.8125 * 10^11, so i think i did good.

breaking it down, this is what was involved.

(4x^-1)^2 is equal to 4^2 * (x^-1)^2 which is equal to 4^2 * x^(-1*2) which is equal to 4^2 * x^(-2)

the exponent arithmetic properties that are involved are:

(x^a)^b is equal to x^(a*b).

(x^a * y^b) ^ c is equal to x^(a*c) * y^(b*c).

the same properties apply to (2x^3)^3.

that is equal to 2^3 * (x^3)^3 which is equal to 2^3 * x^(3*3) which is equal to 2^3 * x^9.

you were then left with:

4^2 * x^(-2) * 2^3 * x^9

the other property of exponents involved is:

x^a * x^b is equal to x^(a + b) and:

x^a / x^b is equal to x^(a-b)

there's another property that states:

x^(-a) is equal to 1/x^a.

your expression of 4^2 * x^(-2) * 2^3 * x^9 becomes:

4^2 * 2^3 * x^9 * x^-2)

simplifying and using the properties of exponents, you get:

4^2 * x^(-2) * 2^3 * x^9 becomes 16 * x^(9-2) * 8 which becomes 128 * x^7.

here's some good references on exponent arithmetic that you might find useful.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut26_exp.htm

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut23_exppart1.htm

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut24_exppart2.htm

let me know if you have any further questions regarding this.