Question: I have the following question:

Is the following statement true or false? 

\*\* All logs to base 2

log2n is a memeber of O(log(n))

My attempt:

log2n - clogn <= 0  

log2 + logn - clogn <= 0
1 + logn(1-c) <= 0

Now correct me if I'm wrong, but I have to find values for n (variable) and c (constant) which either prove or disprove this...

Generally this seems to be true:

Choose n0 = 2, c = 3 -> TRUE
 n1 = 3, c = 3 -> TRUE
 n2 = 4, c = 3 -> TRUE

Therefore, the statement seems true, as logn is increases as n does. But there are also values for which the above statement will never hold:

e.g. 

Choose c = 1 evaluates to greater than zero regardless of the increasing value of n.

So I am confused as to whether this is true or false....

Answer: 
$$\log\_2(2n)=\log\_2(2)+\log\_2(n)\leq \log\_2(n)+\log\_2(n)=2\log\_2(n)$$ for $n\geq 2$.
Let $a>0$ constant. Then:

$\log\_a(n) = \frac{\ln(n)}{\ln(a)} \in \mathcal{O}(\ln(n))$

Therefore, all logarithms share one $\mathcal{O}$-class.