Question 145725
Well, you could begin by recalling that: {{{Log[b](x) = y}}} means {{{b^y = x}}} and that {{{Log[b](x^n) = n*Log[b](x)}}} and that {{{Log[b](b) = 1}}}
Also, {{{0.0001 = 10^(-4)}}}
Now let's see what all of this has to do with your problem:
{{{Log[10](.0001) = Log[10](10^(-4))}}}={{{-4*Log[10](10) = -4(1)}}} = {{{-4}}}