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I'm supposed to prove the theorem below either by definition or the limit test. I understand both of these methods but I'm having a hard time applying them to this particular theorem.

$d > 1, \log_d(n) \in \Theta(\log_2(n))$

I know that by definition, this states $c_1\log_2(n) \le \log_d(n) \le c_2\log_2(n)$. Then you would go about solving the left side and the right side to prove it. But I don't know how to go about that with this problem.

With the limit test, it's like so:

$\lim\limits_{n \to \infty} \frac{\log_d(n)}{\log_2(n)} = 0 \lt c \lt \infty$

Again, I don't know where to go from here.

I think what's throwing me off is that I'm dealing with a more general equation rather than a $log$ with a specific base.

1 Answers 1

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HINT

You may wish to consider the identity $$\frac{ \log_d (n)}{\log_2 (n) }=\frac{ \log_n (2)}{\log_n (d) }=\log_d(2)$$

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    :-D Perfect(ly my answer)2017-01-14