Why is $\log\frac{1}{2} = -\log(2)$ Why does $\log\frac{1}{2} = -\log(2)$
What rule is being used?
EDIT:  Wow, that was fast.  Thanks for the replies.  I saw it shortly after I posted it.
 A: Hint: Apply the rule $\log(a/b) \equiv \log a - \log b$.
A: Hint: $\frac{1}{2}=2^{-1}$ and $\log{a^b}=b\log{a}$
A: Because of the generat property of logarithms
$$\log\frac xy=\log
 x-\log y\implies \log\frac12=\log 1-\log 2=-\log 2$$
A: Recall what $y=\log_a(x)$ means. This means, given $a$ and $x$, $y$ is such that $x=a^y$. Hence, $$\dfrac1x = \dfrac1{a^y} = a^{-y}$$
Hence, $\log_a\left(\dfrac{1}{x}\right) = -y = -\log_a(x)$.
A: All of the above replies are exactly right. Here's a "conceptual" way of looking at it. We want to prove that, in general, $ \log_b \left( \dfrac {1}{x} \right) = - \log_b (x) $. Well, think of it this way. 
Let $ \log_b \left( x \right) = p $. 
Then, $ x = b^p $. That is, we have to raise $b$ to the $p$th power to get $x$. But to get $\dfrac{1}{x}$, we know, from exponential properties, that we have to raise $b$ to the same power but negative of that. Why? Simply because $$ b^{-p} = \dfrac {1}{b^p} $$ by definition. Thus, $ b^{-p} = \dfrac {1}{x} $, so we have our result. 
A: We have $\log ab = \log a + \log b$ and $\log a^b = b \log a$. You should verify these identities using whatever definition of the natural logarithm is at your disposal. Accompanied by the fact that $\log 1 = 0$, the identity in the original post follows.
A: This logarithm rule, $\log \frac{x}{y}=\log x- \log y$, is really from the exponent rule $\frac{x^a}{a^b}=x^{a-b}$. See how they are similar?
A: $\frac{1}{2} \cdot 2=a^{\log_a{\frac{1}{2}}}a^{\log_a{2}}=a^{\log_a{\frac{1}{2}}+\log_a{2}}=a^0$
$\log{\frac{1}{2}}+\log{2}=0$
Therefore:
$\log{\frac{1}{2}}=-\log{2}$
A: $g^{\log_{g}a}=a$ for $g>0\wedge g\neq0\wedge a>0$. This can be
used as definition of $\log_{g}a$. 
Based on that (not using any
rules for logs) you find:
$10^{\log\frac{1}{2}}=\frac{1}{2}=2^{-1}=\left(10^{\log2}\right)^{-1}=10^{-\log2}$
so $\log\frac{1}{2}=-\log2$.
Here $g=10$ and $\log a$ is by convention an abbreviation of $\log_{10}a$.
