$$O(\epsilon^{-2} * log(2/\delta)) = O(\epsilon^{-2} * log(1/\delta)) \,\,\,\,\,\;\;\; for \;0 < \delta < 1 \;and\; 0 < \epsilon < 0.5 $$
is this true? If we apply some logarithm laws to the first Big O we get:
$$O(\epsilon^{-2}*log(2) + \epsilon^{-2} * log(1/\delta))$$
where it would seem to me the equation only holds if $log(1/\delta) \ge log(2) $, but maybe some Big-O-magic makes it true for any $\delta$?
The reason I'm confused about this is, the way I understand Big O, e.g. $O(log(c/\delta))$ for some $c$ would be the same as $O(log(1/\delta))$ , but I don't know if this also holds with multiple parameters.