Law of Products and Quotients

Law of Products

If the base of the power is a product, the exponent should be applied to each factor.
\begin{array}{rcl} (3x^2y^4)^2 & = & 3^2 \cdot (x^2)^2 \cdot (y^4)^2 \\ & = & 9x^4y^8 \end{array}

It is generally easier to apply the Law of Negative Exponents before the Law of Products.\begin{array}{rcl} (2x^{-3}y^2)^{-3} & = & \frac{1}{(2x^{-3}y^2)^3} \\ & = & \frac{1}{8x^{-9}y^6} \\ & = & \frac{x^9}{8y^6} \end{array}

Common Mistake: If the base of a power is a sum, we have no rules! You must use distributive property/”FOIL”.

(a+b)^2 \neq a^2 + b^2

Wrong Answer

\begin{array}{rcl} (a+b)^2 & = & (a+b)(a+b) \\ & = & a^2+ab+ab+b^2 \\ & = & a^2+2ab+b^2 \end{array}

Right Answer

Law of Quotients

If the base of the power is a quotient, the exponent should be applied to the numerator and the denominator.
\Big(\frac{x^2}{y^5}\Big)^3=\frac{x^6}{y^{15}}

It is generally easier to apply the Law of Negative Exponents before the Law of Quotients.\Big(\frac{2x^3}{y^{-4}}\Big)^{-3}=\Big(\frac{y^{-4}}{2x^3}\Big)^3 =\frac{y^{-12}}{8x^9}=\frac{1}{8x^9y^{12}}

Practice

EXAMPLE 1

Simplify then evaluate: 2^{11} \div 2^8

SOLUTION

Using the Law of Division: 2^{11} \div 2^8=2^{11-8}=2^3

Evaluating: 2^3=8

EXAMPLE 2

Simplify: (m^0)^{10}

SOLUTION

Using the Law of Powers: (m^0)^{10}=m^0

We should more properly use the Law of Zero Exponents to write: m^0=1

EXAMPLE 3

Find the value of x that makes the equation true: 4^x \times 4^4=4^7

SOLUTION

Since we are multiplying the two powers, we use the Law of Multiplication: x+4=7

Therefore: x=3

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