1.$log 1$ +$log 2$+$log 3$+...........+$log 100$=? Here the base is 10.
2.In k based log what is the answer?
10 based log
- Abdul Muntakim Rafi
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Re: 10 based log
Part 1: $log_{10} 100!$ The decimal representation is too big.
Please make the second part clear
Please make the second part clear
Last edited by *Mahi* on Fri Apr 22, 2011 6:30 pm, edited 1 time in total.
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Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
- Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
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Re: 10 based log
can u please tell me how you did it?i want an explanation.and the second part asks if there can be found a general method for any base such as 1,2,3,..........,k
Man himself is the master of his fate...
Re: 10 based log
By the general rules of log, $log_k m+ log_k n=log_k mn $
And the second part is the same , $log_k 100!$
And the second part is the same , $log_k 100!$
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
- Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Re: 10 based log
I cannot believe I picked such an easy question.I was looking for a question to post on the forum.I went to a group and posted it from there.But I was such a fool that I didn't think about it.However,thank u for the answer.
Man himself is the master of his fate...
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Re: 10 based log
100! has 24 zero in it's last so $100!=a.10^(24)$. $log100!=loga.10^(24)$ then?