Use the definition of big-Theta to show that log: (9n) = (login) (log base-3 of 9n is big-Theta of log base-2 of n). Follow the following steps - a. First show logs (9n) = O(log2n). b. Then show log3 (9n) = 2(logan). c. (C1, C2, max(no, n'o)) will be the triple you were looking for where (C1, no) and (C2, no) are the pairs returned from the previous two parts. No other steps are needed at this part except to give the triple