How to Solve Powers of Products and Quotients

How to Solve Powers of Products and Quotients

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Discovering a Product’s Power

In order to simplify a product’s power in \(2\) exponential expressions, one can utilize the power of a product rule of exponents. This separates the power of a product of factors into the product of the powers of the factors. For example, look at \((pq)^3\). You start via utilizing the associative and commutative properties of multiplication for regrouping the factors.

\((pq)^3 \ = \ (pq) \ \times \ (pq) \ \times \ (pq) \ = \ p \ \times \ p \ \times \ p \ \times \ q \ \times \ q \ \times \ q \ = \ p^3 \ \times \ q^3\)

So, \((pq)^3 \ = \ p^3 \ \times \ q^3\)

THE POWER OF A PRODUCT RULE OF EXPONENTS

With any actual numbers \(a\) as well as \(b\) and any integer \(n\), the power of a product rule of exponents says:

\((ab)^n \ = \ a^n \ \times \ b^n\)

Discovering a Quotient’s Power

In order to simplify the power of a quotient with \(2\) expressions, one can utilize the power of a quotient rule. It says the power of a quotient of factors equals the quotient of the powers of the factors. For instance, here is an example:
\((e^{-2} \ \times \ f^2)^7 \ = \ \frac{f^{14}}{e^{14}}\)
Now redo the original problem in a different way and see the outcome.
\((e^{-2} \ \times \ f^2)^7 \ = \ (\frac{f^2}{e^2} \ )^7 \ = \ \frac{f^{14}}{e^{14}}\)
Based on the last \(2\) steps, it seems one can utilize the power of a product rule for a power of a quotient rule.
\((e^{-2} \ \times \ f^2)^7 \ = \ (\frac{f^2}{e^2} \ )^7 \ = \ \frac{(f^2)^7}{(e^2)^7} \ = \ \frac{f^{2 \times 7}}{e^{2 \times 7}} \ = \ \frac{f^{14}}{e^{14}}\)

THE POWER OF A QUOTIENT RULE OF EXPONENTS

With any true numbers \(a\) and \(b\) and any integer \(n\), the power of a quotient rule of exponents says that: \((\frac{a}{b} \ )^n \ = \ \frac{a^n}{b^n}\)

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