algorithm - Proof of O(loga n) = O(logb n), for any base a or b -

i revising exams , question appears on past paper:

show o(loga n) = o(logb n) choice of logarithmic bases , b working mathematical definition of order notation f(n) e o(g(n)).

could please show me how solve this?

the rule changing base of logarithm is: log_b(n) = log_a(n) / log_a(b).

this implies log_b(n) = o(log_a(n)) , symmetry log_a(n) = o(log_b(n)).