Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.3 - Logarithmic Functions - Exercise Set - Page 700: 31



Work Step by Step

When the properties of exponents are used, $\dfrac{1}{\sqrt2} = 2^{-\frac{1}{2}}$. Thus, the given expression is equivalent to $\log_2{(2^{-\frac{1}{2}})}$ Let $\log_{2}{(2^{-\frac{1}{2}})} = x$ RECALL: $\log_b{y} =x \longrightarrow b^x=y$ Use the rule above, where $y=2^{-\frac{1}{2}}$ and $b=2$, to obtain: $2^x=2^{-\frac{1}{2}}$ Use the rule "If $a^x=a^y$, then $x=y$" to obtain: $x=-\frac{1}{2}$
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