# Big O - algorithm comparison

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Suppose program A takes (2^n)/1000 units of time and program B takes 1000(n^2) units. For what values of n does program A take less time than program B.

I am really looking for a detailed explanation on this problem - to check my answer.

Thanks.

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Ta = 2^n/1000; Tb = 1000*(n^2)

If Ta < Tb,

2^n/1000 < 1000*(n^2)

(2^n)/n^2 < 1,000,000

Take log on both sides,

n*log 2 - 2log n < 6

n < (6 + 2logn)/log ...

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