Question

Exponent Rule of algebra ll What is the order of the problem in the Picture Please help me-Exponent power rule-Quotient rule-Product rule-Negative exponents rule In order

Answer 1

When an exponent has another exponent on top of it, they can be multiplied.

`(A^2)^3=A^(2\cdot3)=A^6`

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).

`(A^2\cdot A^3)/(A^5)=A^(2+3-5)=A^0`

In the problem, we first appy the exponent power rule, in order to eliminate the parenthesis and have all the factors multypling:

`((2^2x^2y^5)^2)/((2xy^2)^3(3x^3)^2)=(2^4x^4y^(10))/(2^3x^3y^6\cdot3^2x^6)`

Then, to group the similar factors, we apply the product rule in the denominator, as the numerator is alrready simplified:

`(2^4x^4y^(10))/(2^3x^3y^6\cdot3^2x^6)=(2^4x^4y^(10))/(2^3x^(3+6)y^63^2)=(2^4x^4y^(10))/(2^33^2x^9y^6)`

Now, we apply the quotient rule for the factors with the same base:

`(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`

Finally, the negative exponents rule to put the numbers with negative exponents in the denominator with positive exponent:

`2\cdot3^(-2)\cdot x^(-5)\cdot y^4=(2y^4)/(3^2x^5)`

Order:

Expanded power rule

Product rule

Quotient rule

Negative exponent rule