College Algebra (6th Edition)

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

Chapter 7 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-7) - Page 7074: 17


$3log_{5}(x) + \frac{1}{2}log_{5}(y) - 3$

Work Step by Step

$$log_{5}(\frac{x^{3}\sqrt {y}}{125})$$ Taking care of the exponents seems like a good first step: $log_{5}(\frac{x^{3}\sqrt {y}}{125}) = log_{5}(\frac{x^{3}y^{\frac{1}{2}}}{5^{3}}) = log_{5}(5^{-3}x^{3}y^{\frac{1}{2}})$ All three terms are being multiplied, so by the Rules of Logarithms, we can expand as follows: $log_{5}(5^{-3}x^{3}y^{\frac{1}{2}}) = log_{5}(5^{-3}) + log_{5}(x^{3}) + log_{5}(y^{\frac{1}{2}})$ Since each logarithmic term has an associated exponent, we can use the Rules of Logarithms to justify bringing them to the front of the term as constants: $ log_{5}(5^{-3}) + log_{5}(x^{3}) + log_{5}(y^{\frac{1}{2}}) = -3log_{5}(5) + 3log_{5}(x) + \frac{1}{2}log_{5}(y)$ Finally, when the term of a logarithm is the same as its base, the Rules of Logarithms justify the term equals to 1: $ -3(log_{5}(5)) + 3log_{5}(x) + \frac{1}{2}log_{5}(y) = -3(1) + 3log_{5}(x) + \frac{1}{2}log_{5}(y)$ Therefore, the final expansion of the original logarithm is: $$3log_{5}(x) + \frac{1}{2}log_{5}(y) - 3$$
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