Simplifying Logarithmic Expressions

Since $b^1=b$

${\log }_{b}1=0$

for any $b$

Since $b^0=1$

${\log }_{b}0$ is undefined

for all $b$

Since there is no $x$ for which $b^x=0$

${\log }_{b}x$ is undefined

if $x$ is negative

It may seem like ${\log }_{-2}\left(-8\right)$ should equal $3$ , since ${\left(-2\right)}^3=-8$ .

But on the other hand, ${\log }_{-2}\left(-4\right)$ doesn't mean anything: the equation ${\left(-2\right)}^x=-4$ has no solution.

${\log }_{b}xy={\log }_{b}x+{\log }_{b}y$

Since $b^m\cdot b^n=b^{m+n}$

${\log }_b\frac{x}{y}={\log }_{b}x-{\log }_{b}y$