SOLUTION: WHAT IS THE (X^4Y^8)^1/2*(X^1/2)6

Question 85206This question is from textbook ALGEBRA 1
: WHAT IS THE (X^4Y^8)^1/2*(X^1/2)6 This question is from textbook ALGEBRA 1

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given:
.
(X^4Y^8)^1/2*(X^1/2)6
.
I presume that the 6 at the very end of the problem is to be an exponent so that the
problem really is:
.
(X^4Y^8)^1/2*(X^1/2)^6
.
The exponent of 1/2 for the first (left hand) set of parentheses tells you to take the square
root of the quantity inside the parentheses. You can also get this through using the power rule
for exponents and it tells you for a problem such as (x^4*y^8)^1/2 to multiply the 1/2 times
each of the exponents on the factors inside the parentheses. So you multiply 1/2 times 4
to get an answer of 2 as the exponent of x and you multiply the 1/2 times 8 to get 4 as
the exponent of y. The result is that you can replace (x^4*y^8)^1/2 by (x^2*y^4).
When you make that replacement, the problem is reduced to:
.
(x^2*y^4)*(x^1/2)^6
.
Next you can apply the same power rule to the last factor of (x^1/2)^6. To apply this rule
simply multiply 6 times the exponent inside the parentheses. The product of 6 times 1/2
is 3 which then becomes the exponent of x to make the term (x^1/2)^6 equivalent to x^3. So
make this substitution in the reduced form of the problem and the new reduced form becomes:
.
(x^2*y^4)*(x^3)
.
You can then multiply x^2 by x^3 to get x^5 (add the exponents) and the final answer is
then x^5*y^4.
.
Hope this shows you how to do problems of this sort through using the power rule for exponents
and the rule for multiplying two like quantities by adding their exponents.

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