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Q. 101

Expert-verified

Elementary Algebra

Found in: Page 710

Book edition 2nd

Author(s) Lynn Marecek, MaryAnne Anthony-Smith

Pages 1240 pages

ISBN 9780998625713

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Short Answer

In the following exercises, simplify each expression using the Product Property for Exponents.
$q^{27} \cdot q^{15}$

The value of $q^{27} \cdot q^{15} is q^{42} .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

In the given exercises, we have to simplify given expression using the Product Property for Exponents.
$q^{27} \cdot q^{15}$

Step 2. Use the product property, am·an=am+n

$q^{27} \cdot q^{15} = q^{27 + 15}$

Simplify.

$q^{27} \cdot q^{15} = q^{42}$

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