Thread: Proving big omega second question

1. Proving big omega second question

Hi,

I'm looking for help in proving $\log_2 n - \log_{10} n \in \Omega(\log_{12} n)$. I'm having a really hard time with determining this, especially since its not a polynomial.

Also, the expansion of $\log_2 n - \log_{10} n \in \Omega(\log_{12} n)$ is $\exists c, n_0 \in \mathbb{R}^+, \forall n \in \mathbb{N}, n \geq n_0 \Rightarrow \log_2 n - \log_{10} n \geq c \cdot \log_{12} n$

I tried playing around logarithm rules, but nothing really came out of that for me.

2. Re: Proving big omega second question

$$\log_2 n - \log_{10} n = \dfrac{\ln n}{\ln 2} - \dfrac{\ln n}{\ln 10} = \ln n\left( \dfrac{\ln 10 - \ln 2}{(\ln 10)(\ln 2)} \right) = \ln n \left( \dfrac{\ln 5}{(\ln 10)(\ln 2)} \right)$$

And

$$\log_{12} n = \dfrac{\ln n}{\ln 12}$$

So, you have absolute equality when

$$c = \dfrac{(\ln 5)(\ln 12)}{(\ln 10)(\ln 2)} \approx 2.5>0$$

3. Re: Proving big omega second question

amazing. thanks again

4. Re: Proving big omega second question

Also, can you let me know if there is a way to edit the original post I made? I'm just curious if this is an option.

5. Re: Proving big omega second question

Originally Posted by SlipEternal
$$\log_2 n - \log_{10} n = \dfrac{\ln n}{\ln 2} - \dfrac{\ln n}{\ln 10} = \ln n\left( \dfrac{\ln 10 - \ln 2}{(\ln 10)(\ln 2)} \right) = \ln n \left( \dfrac{\ln 5}{(\ln 10)(\ln 2)} \right)$$

And

$$\log_{12} n = \dfrac{\ln n}{\ln 12}$$

So, you have absolute equality when

$$c = \dfrac{(\ln 5)(\ln 12)}{(\ln 10)(\ln 2)} \approx 2.5>0$$

Could you explain the in between steps of $\ln{n} \cdot (\frac{\ln{10} - \ln{2}}{\ln{10}\cdot \ln{2}}) = \ln{n}(\frac{\ln{5}}{(\ln{10})\cdot(\ln{2})})$?

6. Re: Proving big omega second question

Originally Posted by otownsend
Could you explain the in between steps of $\ln{n} \cdot (\frac{\ln{10} - \ln{2}}{\ln{10}\cdot \ln{2}}) = \ln{n}(\frac{\ln{5}}{(\ln{10})\cdot(\ln{2})})$?
$$\ln a-\ln b=\ln \dfrac a b$$