Stumped by this Exponent Problem?

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3k^(-2) / k^4 = (3 / k^2) / k^4 = 3/ k^6

so
[(3K^-2 / K^4) ^-1] = (3/k^6)^(-1) = k^6/3

so
3K^-2 / K^4) ^-1] * 2 / k = k^6/3 * 2/k
--- = 2(k^5) / 3

a million. (a million/a-a million/b) / (a million/a+a million/b) = ((b-a)/ab) / ((b+a)/ab) = b-a/ab * ab/b+a =b-a/b+a 2. (a million/d + a million/e)(a million/(d+e)) ((e+d)/ed)*(a million/(d+e) a million/ed 3. (25y^3/x^5) / (a million/(5x^2y)^3 25y^3/x^5 * 125x^6y^3 25(one hundred twenty 5)xy^6 3125xy^6

The original expression=
(2/k)[k^4/(3k^-2)]=
(2/k)[(k^4)(k^2)/3]=
2[k^5/3]=
2k^5/3

[3k^(-2) / K^4]^(-1) * 2 / k =
[3 * k^(-2) / K^4]^(-1) * 2 / k =

Remember that x^a / x^b = x^(a - b).
[3 * k^(-2 - 4)]^(-1) * 2 / k =
[3 * k^(-6)]^(-1) * 2 / k =

Remember that (x * y)^a = x^a * y^a.
3^(-1) * [k^(-6)]^(-1) * 2 / k =

Remember that x^(-a) = 1 / x^a.
(1 / 3^1) * [k^(-6)]^(-1) * 2 / k =
(1 / 3) * [k^(-6)]^(-1) * 2 / k =

Remember that (x^a)^b = x^(a * b).
(1 / 3) * k^(-6 * -1) * 2 / k =
(1 / 3) * k^6 * 2 / k =

Group.
[(1 / 3) * 2] * (k^6 / k) =
(2 / 3) * (k^6 / k) =

Remember that x^a / x^b = x^(a - b).
(2 / 3) * k^(6 - 1) =
(2 / 3) * k^5 =
(2k^5 / 3)

ANSWER: (2k^5 / 3)

Remember these rules when you're working with exponents.

x^(-a) = 1 / xª
xª * x^b = x^(a + b)
xª / x^b = x^(a - b)
(xª)^b = x^(a * b)
(xy)ª = (x * y)ª = xª * yª
√a = a^(½)
ª√x = x^(1 / a)
(x / y)ª = xª / yª
√(x * y) = √x * √y
ª√(x * y) = ª√x * ª√y

I got it. that whole fraction that is raised to the -1 becomes (k to the sixth power) over (3).

k^6 in numerator, k in denominator = k^5

the 2 stays up there

the 3 is in the denominator

(2k^5) / 3

[[3K^-2] / (K^4) ] ^ -1 = (3 / [K^6] ) ^ -1 = (K^6 ) / 3.

[ (K^6 ) / 3 ](2)(1/K) = (2k^5) / 3


Touch left point right.

2[(3k^-2/k^4)^-1]/k
= 2{[3k^(-2 - 4)]^-1}/k
= 2{[3k^-6]^-1}/k
= 2[3^-1k^6]/k
= 2[k^6/3]/k
= (2k^6/3)/k
= (2k^6/3)(1/k)
= (2k^5)/3

First flip the fraction to get rid of the exponent -1.
[(k^4/3k^-2]*2/k
Now move k^-2 into the numerator and make the exponent +
(k^4k^2/3)*2/3k
simplify
(k^6)*2/3k
2k^5/3

[(3K^-2 / K^4) ^-1] * 2 / K
[(3K^-2 * K^-4) ^-1] *2 / K <-- rewrite 1 / K^4 as K^-4
[(3K^-6) ^ -1] * 2 / K <-- combine K^-2 and K^-4
[K^6 / 3] * 2 / K <-- when you raise it to -1 power, you get rid of the negative on the K's exponent, and the 3 gets flipped
2 * K^5 / 3 <-- simplify