How is $O(cn \log_{3/2}n) = O(n \lg{n})$?

Question

I am reading Introduction to Algorithms 3rd edition and it says that $O(cn \log_{3/2}n) = O(n \lg{n})$. But isn't $\lg{x}$ tighter than $\log_{3/2}x$?

Answer 1

Recall that $\text{log}_{2}(n) = \dfrac{\log_{3}(n)}{\log_{3}(2)}$, per the change of base formula. The $\log_{3}(2)$ term is absorbed by the constant factor, which you have from the definition of Big-O.

Answer 2

Hint: $\log_{3/2}n=\ln n/\ln(3/2)$.