College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3: 99

Answer

$27y^\frac{2}{3}$

Work Step by Step

$$\frac{(3y^\frac{1}{4})^3}{y^\frac{1}{12}}$$ When an exponential expression is raised to a power, multiply the exponents: $$=\frac{27y^{(\frac{1}{4} \times 3)}}{y^\frac{1}{12}}$$ $$=\frac{27y^\frac{3}{4}}{y^\frac{1}{12}}$$ When exponential expressions with the same base are divided, subtract the exponent of the denominator from the exponent of the numerator: $$=27y^{(\frac{3}{4}-\frac{1}{12})}$$ To subtract fractions, you need a common denominator; in this case, 12: $$=27y^{(\frac{9}{12}-\frac{1}{12})}$$ $$=27y^{\frac{8}{12}}$$ Simplify all fractions: $$=27y^\frac{2}{3}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.