The distributive property of exponents over multiplication says that when you have multiple factors raised to an exponent, the exponent can be distributed to each of the factors individually.

\((ab)^{x}=a^{x}b^{x}\)

This property can also be applied to division and it can be used in reverse to multiply or divide powers with the same exponent.

Most math teachers call the distributive property of exponents some variation of...

- Product Rule of Exponents (Same Exponent, Different Bases)
- Multiplying Exponents with Different Bases
- Multiplying Powers with Same Exponents

I call it the "distributive property of exponents over multiplication" because the relationship between exponents and multiplication is very similar to the relationship between multiplication and addition.

The relationship between multiplication and addition is called the "distributive property of multiplication over addition" and you can read more about it here.

\(3(2x+5y)=6x+15y\)

\((x^{5}y^{4})^{2}=x^{10}y^{8}\)

There are two different distributive properties.

- The first one (multiplication over addition) is the one that everyone thinks of when they hear "distributive property".
- The second one (exponents over multiplication) is a term I made up to describe one of the exponent rules.

There are a lot of similarities between these two properties.

Both distributive properties have parentheses.

- The distributive property of multiplication over addition has
**addition**inside the parentheses. Each term inside the parentheses is**multiplied**by the outside number. - The distributive property of exponents over multiplication has
**multiplication**inside the parentheses. Each factor inside the parentheses is raised to the outside**exponent**.

When you look at WHY they work, there are also a lot of similarities...

\(3(2x+5y)=6x+15y\)

\(2x+5y\) is the same thing as \(x+x+y+y+y+y+y\).

When you make 3 copies of that, you get...

\(x+x+y+y+y+y+y+\)

\(x+x+y+y+y+y+y+\)

\(x+x+y+y+y+y+y\)

There are a total of 6 x's and 15 y's being added together, so that is why...

\(3(2x+5y)=6x+15y\)

\((x^{5}y^{4})^{2}=x^{10}y^{8}\)

\(x^{5}y^{4}\) is the same thing as \(xxxxxyyyy\).

When you multiply that by itself 2 times, you get...

\((xxxxxyyyy)(xxxxxyyyy)\)

There's a total of 10 x's and 8 y's multiplied together, so that is why...

\((x^{5}y^{4})^{2}=x^{10}y^{8}\)

Exponents can NOT be distributed over addition or subtraction.

\((x+y)^2\neq x^2+y^2\)

If you have an exponent outside of parentheses with addition or subtraction inside, then you need to follow the rules for multiplying polynomials.

Distributive Property of Exponents

\((ab)^{x}=a^{x}b^{x}\)

\((3x)^{4}=81x^{4}\)

\((ab)^{5}=a^{5}b^{5}\)

\((2y)^{5}=32y^{5}\)

- Identify the factors inside the parentheses.
- Distribute the exponent outside the parentheses to each of the factors inside the parentheses.
- Simplify the powers (using the power rule of exponents, if needed).

Simplify \((3x^{4}y)^2\)

The factors in this expression are 3, \(x^4\), and y.

I need to distribute the outside exponent of 2 to each factor.

\((3x^{4}y)^2=3^{2}(x^4)^{2}y^{2}\)

Next, I'll simplify each factor, if possible.

\(3^{2} = 3\times3=9\)

\((x^4)^{2}=x^8\) because of the power rule of exponents.

\(y^2\) cannot be simplified further.

\(3^{2}(x^4)^{2}y^{2}=9x^{8}y^{2}\)

Answer: \((3x^{4}y)^2=9x^{8}y^{2}\)

\((\frac{a}{b})^{x}=\frac{a^{x}}{b^{x}}\)

\((\frac{a}{b})^{7}=\frac{a^{7}}{b^{7}}\)

\((\frac{x^2}{4})^{3}=\frac{x^{6}}{64}\)

\((\frac{5}{xyz})^{3}=\frac{125}{x^{3}y^{3}z^{3}}\)

- Identify all the factors in the numerator.
- Identify all the factors in the denominator.
- Distribute the exponent outside the parentheses to each of the factors in the numerator and the denominator.
- Simplify the powers (using the power rule of exponents, if needed).

Simplify \((\frac{3xy}{7z^4})^{2}\)

The factors in the numerator are 3, x, and y.

The factors in the denominator are 7 and \(z^4\).

I need to distribute the outside exponent of 2 to each factor.

\((\frac{3xy}{7z^4})^{2}=\frac{3^2x^2y^2}{7^2(z^4)^2}\)

Next, I'll simplify each factor, if possible.

\(3^{2} = 3\times3=9\)

\(x^2\) and \(y^2\) cannot be simplified further.

\(7^{2} = 7\times7=49\)

\((x^4)^{2}=x^8\) because of the power rule of exponents.

So...

\(\frac{3^2x^2y^2}{7^2(z^4)^2}=\frac{9x^2y^2}{49z^8}\)

Answer: \((\frac{3xy}{7z^4})^{2}=\frac{9x^2y^2}{49z^8}\)

\(a^{x}b^{x}=(ab)^{x}\)

\(2^{3}4^{3}=8^{3}\)

\(7^{2}x^{2}=(7x)^{2}\)

\(a^{5}b^{5}=(ab)^{5}\)

- Make sure ALL the powers you are multiplying have the
**same exponent**. - Multiply the bases and write the power with one exponent.
- Simplify, if needed.

Simplify \((6^3)(5^3)\).

These two powers do have the same exponent (3).

So, I will multiply the bases together.

\((6^3)(5^3)=30^3\)

There are no variables so I can evaluate this expression.

\(30^{3} = 30\times30\times30=27000\)

Answer: \((6^3)(5^3)=27000\)

\(\frac{a^{x}}{b^{x}}=(\frac{a}{b})^{x}\)

\(\frac{a^{9}}{b^{9}}=(\frac{a}{b})^{9}\)

\(\frac{4^{2}}{x^{2}}=(\frac{4}{x})^{2}\)

\(\frac{5^3x^{3}}{y^{3}}=(\frac{5x}{y})^{3}\)

- Make sure ALL the powers you are dividing have the
**same exponent**. - Divide the bases and write the power with one exponent.
- Simplify, if needed.

Simplify \(\frac{8^{3}}{2^{3}}\).

These two powers do have the same exponent (3).

So, I will divide the bases and rewrite the power.

\(\frac{8^{3}}{2^{3}}=4^3\)

There are no variables so I can evaluate the expression.

\(4^{3} = 4\times4\times4=64\)

Answer:\(\frac{8^{3}}{2^{3}}=64\)

Khan Academy - Powers of products & quotients (structured practice)

Khan Academy - Powers of products & quotients (integer exponents)

Khan Academy - Powers of products & quotients

Math Aids - Products to a Power Worksheet Generator

Math Aids - Quotients to a Power Worksheet Generator

Math Aids - Products and Quotients to a Power Worksheet Generator

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