Answer :
To solve the problem of finding which two monomials can be multiplied together to get the same value as [tex]\((x^2 y^3)^4\)[/tex], let's go through the steps together:
1. First, simplify [tex]\((x^2 y^3)^4\)[/tex]:
This expression can be rewritten using the properties of exponents. When raising a monomial to a power, we multiply the exponents of each variable by that power.
[tex]\[ (x^2 y^3)^4 = x^{2 \cdot 4} y^{3 \cdot 4} = x^8 y^{12} \][/tex]
So, [tex]\((x^2 y^3)^4 = x^8 y^{12}\)[/tex].
2. Now, we need to find which pair of monomials, when multiplied, result in [tex]\(x^8 y^{12}\)[/tex]:
a. Try [tex]\(x y^6\)[/tex] and [tex]\(x y^\beta\)[/tex]:
[tex]\[ \text{Multiplication: } (x y^6) \cdot (x y^\beta) = x^{1+1} y^{6+\beta} = x^2 y^{6 + \beta} \][/tex]
This cannot be equal to [tex]\(x^8 y^{12}\)[/tex] because [tex]\(x^2\)[/tex] doesn't match [tex]\(x^8\)[/tex] and [tex]\(6 + \beta\)[/tex] would have to be 12, but that leaves [tex]\(\beta\)[/tex] as 6, making only the [tex]\(y\)[/tex] part correct without addressing the [tex]\(x\)[/tex] part properly.
b. Try [tex]\(x^2 y^3\)[/tex] and [tex]\(x^4 y^4\)[/tex]:
[tex]\[ \text{Multiplication: } (x^2 y^3) \cdot (x^4 y^4) = x^{2+4} y^{3+4} = x^6 y^7 \][/tex]
This result does not match [tex]\(x^8 y^{12}\)[/tex].
c. Try [tex]\(x^2 y^2\)[/tex] and [tex]\(x^6 y\)[/tex]:
[tex]\[ \text{Multiplication: } (x^2 y^2) \cdot (x^6 y) = x^{2+6} y^{2+1} = x^8 y^3 \][/tex]
This result does not match [tex]\(x^8 y^{12}\)[/tex].
d. Try [tex]\(x^2 y^3\)[/tex] and [tex]\(x^6 y^9\)[/tex]:
[tex]\[ \text{Multiplication: } (x^2 y^3) \cdot (x^6 y^9) = x^{2+6} y^{3+9} = x^8 y^{12} \][/tex]
This result matches [tex]\(x^8 y^{12}\)[/tex].
So, the correct pair of monomials that can be multiplied together to get [tex]\((x^2 y^3)^4 = x^8 y^{12}\)[/tex] is:
[tex]\[ \boxed{x^2 y^3 \text{ and } x^6 y^9} \][/tex]
1. First, simplify [tex]\((x^2 y^3)^4\)[/tex]:
This expression can be rewritten using the properties of exponents. When raising a monomial to a power, we multiply the exponents of each variable by that power.
[tex]\[ (x^2 y^3)^4 = x^{2 \cdot 4} y^{3 \cdot 4} = x^8 y^{12} \][/tex]
So, [tex]\((x^2 y^3)^4 = x^8 y^{12}\)[/tex].
2. Now, we need to find which pair of monomials, when multiplied, result in [tex]\(x^8 y^{12}\)[/tex]:
a. Try [tex]\(x y^6\)[/tex] and [tex]\(x y^\beta\)[/tex]:
[tex]\[ \text{Multiplication: } (x y^6) \cdot (x y^\beta) = x^{1+1} y^{6+\beta} = x^2 y^{6 + \beta} \][/tex]
This cannot be equal to [tex]\(x^8 y^{12}\)[/tex] because [tex]\(x^2\)[/tex] doesn't match [tex]\(x^8\)[/tex] and [tex]\(6 + \beta\)[/tex] would have to be 12, but that leaves [tex]\(\beta\)[/tex] as 6, making only the [tex]\(y\)[/tex] part correct without addressing the [tex]\(x\)[/tex] part properly.
b. Try [tex]\(x^2 y^3\)[/tex] and [tex]\(x^4 y^4\)[/tex]:
[tex]\[ \text{Multiplication: } (x^2 y^3) \cdot (x^4 y^4) = x^{2+4} y^{3+4} = x^6 y^7 \][/tex]
This result does not match [tex]\(x^8 y^{12}\)[/tex].
c. Try [tex]\(x^2 y^2\)[/tex] and [tex]\(x^6 y\)[/tex]:
[tex]\[ \text{Multiplication: } (x^2 y^2) \cdot (x^6 y) = x^{2+6} y^{2+1} = x^8 y^3 \][/tex]
This result does not match [tex]\(x^8 y^{12}\)[/tex].
d. Try [tex]\(x^2 y^3\)[/tex] and [tex]\(x^6 y^9\)[/tex]:
[tex]\[ \text{Multiplication: } (x^2 y^3) \cdot (x^6 y^9) = x^{2+6} y^{3+9} = x^8 y^{12} \][/tex]
This result matches [tex]\(x^8 y^{12}\)[/tex].
So, the correct pair of monomials that can be multiplied together to get [tex]\((x^2 y^3)^4 = x^8 y^{12}\)[/tex] is:
[tex]\[ \boxed{x^2 y^3 \text{ and } x^6 y^9} \][/tex]
