When an exponent has another exponent on top of it, they can be multiplied.
[tex](A^2)^3=A^{2\cdot3}=A^6[/tex]
When two numbers with the same base are factors in a multiplication, their exponents can be added. The same for a quotient, but with the difference that the exponents are substracted (the number in the denominator is, in fact, the base with the negative exponent).
[tex]\frac{A^2\cdot A^3}{A^5}=A^{2+3-5}=A^0[/tex]
In the problem, we first appy the exponent power rule, in order to eliminate the parenthesis and have all the factors multypling:
[tex]\frac{(2^2x^2y^5)^2}{(2xy^2)^3(3x^3)^2}=\frac{2^4x^4y^{10}}{2^3x^3y^6\cdot3^2x^6}[/tex]
Then, to group the similar factors, we apply the product rule in the denominator, as the numerator is alrready simplified:
[tex]\frac{2^4x^4y^{10}}{2^3x^3y^6\cdot3^2x^6}=\frac{2^4x^4y^{10}}{2^3x^{3+6}y^63^2}=\frac{2^4x^4y^{10}}{2^33^2x^9y^6}[/tex]
Now, we apply the quotient rule for the factors with the same base:
[tex]\frac{2^4x^4y^{10}}{2^33^2x^9y^6}=2^{4-3}\cdot3^{-2}\cdot x^{4-9}\cdot y^{10-6}=2\cdot3^{-2}\cdot x^{-5}\cdot y^4[/tex]
Finally, the negative exponents rule to put the numbers with negative exponents in the denominator with positive exponent:
[tex]2\cdot3^{-2}\cdot x^{-5}\cdot y^4=\frac{2y^4}{3^2x^5}[/tex]
Order:
Expanded power rule
Product rule
Quotient rule
Negative exponent rule
