Question 386744
{{{2^(3n-1)>=(1/8)^n}}}.  This inequality is the same as {{{2^(3n-1)>=2^(-3n)}}}.  Now take the base-2 logarithm of both sides of the inequality, to get 
{{{3n - 1 >= -3n}}}.  This inequality is true because we know the base-2 logarithm function is an INCREASING function.  Hence, {{{6n >= 1}}}, or {{{n >= 1/6}}}.