[Solved]: Does ln n ∈ Θ(log2 n)?

Problem Detail: 

Is that statement false or true? I believe it's false because ln(n) = log base e of n. So therefore, log base 2 of n can be a minimum because in 2^x = n, x will always be less than y in e^y = n. However can it ever be proven that log base 2 of n can be a maximum?

Asked By : Jonathan

Answered By : Luke Mathieson

Remember your log laws:

$$ \log_{a}b = \frac{log_{x}b}{\log_{x}a} $$

So $$ \ln n = \frac{\log_{2}n}{\log_{2}e} $$

Given this, can you think of three constants $c_{1}$, $c_{2}$ and $n_{0}$ such that $\ln n \leq c_{1}\cdot\log_{2} n$ and $\ln n \geq c_{2}\cdot \log_{2}n$ for all $n \geq n_{0}$?

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/48006