College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.3 - Logarithmic Functions - 4.3 Exercises - Page 388: 34


$a.\displaystyle \qquad \frac{1}{4}$ $b.\displaystyle \qquad -\frac{1}{2}$ $c.\displaystyle \qquad \frac{3}{2}$

Work Step by Step

By definition, $\log_{a}x=y \Leftrightarrow a^{y}=x$ ($\log_{a}x$ is the exponent to which the base $a$ must be raised to give $x$.) What the definition states is that $\log_{a}(x)$ and $a^{y}$ are inverse functions, $\log_{a}(a^{x})=x$ and $a^{\log_{a}(x)}=x$ "ln" is special annotation for the natural logarithm, $\log_{e}$ (with base e). "log" (without a base) stands for $\log_{10}$, the common logarithm. --- $a.$ $2^{2}=4$, so $2=4^{1/2}.$ $\sqrt{2}=2^{1/2}=(4^{1/2})^{1/2}=4^{1/4}$ Therefore, $\displaystyle \log_{4}\sqrt{2}=\log_{4}4^{1/4}=\frac{1}{4}$ $b.$ $2^{2}=4$, so $2=4^{1/2}.$ $\displaystyle \frac{1}{2}=2^{-1}=(4^{1/2})^{-1}=4^{-1/2}$ Therefore, $\displaystyle \log_{4}\frac{1}{2}=\log_{4}4^{-1/2}=-\frac{1}{2}$ $c.$ $2^{2}=4$, so $2=4^{1/2}.$ $8=2^{3}=(4^{1/2})^{3}=4^{3/2}$ Therefore, $\displaystyle \log_{4}8=\log_{4}4^{3/2}=\frac{3}{2}$
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