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