Arrange the following functions in increasing asymptotic order: A. n1/3 B. en…

Final increasing asymptotic order: n^(1/3), n·log_9(n), n^(7/4), 1.0000001^n, e^n

• log_9(n) = (ln n)/(ln 9), so n·log_9(n) = Θ(n log n).

• Compare polynomials: for exponents a<b, n^a = o(n^b). Thus n^(1/3) = o(n^(7/4)).

• Compare n·log n and n^(7/4): n·log n = n·(log n) and n^(7/4) = n·n^(3/4). Since n^(3/4) grows faster than log n, n·log n = o(n^(7/4)).

• Exponentials dominate polynomials: for any constant a>1, a^n = ω(n^k) for every fixed k. Therefore 1.0000001^n outgrows n^(7/4).

• Among exponentials, larger base → faster growth, so 1.0000001^n = o(e^n).

Combining these facts gives the increasing order: n^(1/3) < n·log_9(n) < n^(7/4) < 1.0000001^n < e^n.