# How do you evaluate log_4 (1/2)?

Jan 7, 2016

${\log}_{4} \left(\frac{1}{2}\right) = - \frac{1}{2}$

#### Explanation:

Using the following properties:

• ${\log}_{a} \left(a\right) = 1$

• $\log \left({a}^{x}\right) = x \log \left(a\right)$

we have

${\log}_{4} \left(\frac{1}{2}\right) = {\log}_{4} \left({4}^{- \frac{1}{2}}\right)$

$= - \frac{1}{2} {\log}_{4} \left(4\right)$

$= - \frac{1}{2} \left(1\right)$

$= - \frac{1}{2}$

Jan 7, 2016

${\log}_{4} \left(\frac{1}{2}\right) = \left(- \frac{1}{2}\right)$

#### Explanation:

Based on the meaning of $\log$ and exponents:
If
$\textcolor{w h i t e}{\text{XXX}} {\log}_{4} \left(\frac{1}{2}\right) = c$
then
$\textcolor{w h i t e}{\text{XXX}} {4}^{c} = \frac{1}{2}$

We know that
$\textcolor{w h i t e}{\text{XXX}} {4}^{c} \ge 1$ for $c \ge 0$
So
$\textcolor{w h i t e}{\text{XXX}} {4}^{c} = \frac{1}{2} \rightarrow c < 0$

Also
$\textcolor{w h i t e}{\text{XXX}} {4}^{\left(- k\right)} = \frac{1}{{4}^{k}}$

Which leads us to asking for what value of $k$ does
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{{4}^{k}} = \frac{1}{2}$?

Since
$\textcolor{w h i t e}{\text{XXX}} {4}^{\frac{1}{2}} = \sqrt{4} = 2$

$\frac{1}{2} = \frac{1}{{4}^{\frac{1}{2}}} = {4}^{- \frac{1}{2}}$
and
$\textcolor{w h i t e}{\text{XXX}} c = \left(- \frac{1}{2}\right)$