Question 184407
{{{.01 = 10^(-2)}}}
{{{log(.01) = log(10^(-2))}}}
From the general rule:
{{{log(a,b^c) = c*log(a,b)}}}
{{{log(10^(-2)) = (-2)*log(10)}}}
Note that 
{{{log(10) = 1}}}
{{{log(10^(-2)) = (-2)*1}}}
{{{log(.01) = -2}}} answer