Expert: Sherman D. - 8/14/2008

Question
QUESTION: a. the square root of 8x^5y^4

b. the cube root of 4x^2y^4 times the square root of 2xy

c. the square root of the square root of x divided by the cube root of x^5

d. (  x ^ -1/2  ) ^-3
    __________
    
    x ^ -2/3

e. the 5th root of x square root of x^3y

ANSWER: a.)
sqrt(8x^5y^4)
sqrt((4x^4y^4)(2x))

just sqrt the 4 and divide the exponents of 4 by 2

4x^2y^2sqrt(2x)

---------------------------------

b.)
cbrt(4x^2y^4) * sqrt(2xy)

this is the same as saying

(4x^2y^4)^(1/3) * (2xy)^(1/2)
(2^2)^(1/3)x^(2/3)y^(4/3) * 2^(1/2)x^(1/2)y^(1/2)

(2^(2/3) * 2^(1/2)) * (x^(2/3) * x^(1/2)) * (y^(4/3) * y^(1/2))

now just add the exponents

2^((2/3) + (1/2)) * x^((2/3) + (1/2)) * y^((4/3) + (1/2))
2^((4 + 3)/6) * x^((4 + 3)/6) * y^((8 + 3)/6)
2^(7/6) * x^(7/6) * y^(11/6)
2(2^(1/6)) * x(x^(1/6) * y(y^(5/6))

ANS : (2xy)(2xy^5)^(1/6) thats what i got, but just to be sure, recheck it.

-------------------------------

c.)
sqrt(sqrt(x))/cbrt(x^5)
((x^(1/2))^(1/2))/((x^(5))^(1/3))
(x^(1/4))/(x^(5/3))
x^((1/4) - (5/3))
x^((3 - 20)/12)
x^(-17/20) or 1/(x^(17/20))

---------- FOLLOW-UP ----------

QUESTION: So what would i do for the following:

d. (  x ^ -1/2  ) ^-3
   __________
   
   x ^ -2/3

e. the 5th root of x square root of x^3y

Thank you soo much

Answer
d.)
(x^(-1/2))^(-3) / x^(-2/3)

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

since x^(-2/3) has a negative exponent and is on the bottom, it becomes

x^(3/2) * x^(2/3)

x^((3/2) + (2/3))

x^((9 + 4)/6)

x^(13/6)

----------------------------------

if by this you mean

5thrt(x) * sqrt(x^3y)

x^(1/5) * (x^3y)^(1/2)

x^(1/5) * (x^3)^(1/2) * y^(1/2)

x^(1/5) * x^(3/2) * y^(1/2)

x^((1/5) + (3/2)) * y^(1/2)

x^((2 + 15)/10) + y^(1/2)

x^(17/10)y^(1/2)