Answer
$a.\displaystyle \qquad \frac{1}{4}$
$b.\displaystyle \qquad -\frac{1}{2}$
$c.\displaystyle \qquad \frac{3}{2}$
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.
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$a.$
$2^{2}=4$, 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,
$\displaystyle \log_{4}8=\log_{4}4^{3/2}=\frac{3}{2}$
