# how can i evaluate logarithmic series

If I have $$\log_{21} 1 + \log_{21}3+\log_{21} 7+...+\log_{21} 441$$ and all the values are divisors of 441. How would i evaluate this sum?

I know the multiplication rule would make it $$\log_{21} (1*3*7*...*441)$$ but i don't know how to evaluate this.

I can raise it to the 21st power. but I still don't know how to evaluate the product

• If you are lucky $1 \times 3 \times 7 \times 9 \times 21 \times 49 \times 63 \times 147 \times 441=21^n$ for some integer $n$. Can you find $n$? – Henry May 13 at 23:12

You just have to find the number of factors of $$441$$.
These factors are: $$1, 3, 7, \text{... }, \frac{441}7, \frac{441}3, \frac{441}1$$
Also notice that $$\log_{21} {1} + \log_{21} {441} = \log_{21} {3} + \log_{21}{\frac{441}3} = \log_{21} {7} + \log_{21}{\frac{441}7} = \text{ ... }= 2$$
• No; All of these "pairs" are equal to $2$. So, the answer would have to be the number of pairs times $2$. – minori minus May 13 at 23:22
• Also note that only $21$ doesn't have a pair. – minori minus May 13 at 23:23
• I mean pairs of factors: 1 and 441, 3 and 147, 7 and 63, 9 and 41. There are 4 pairs, which you can see as $\log_{21} 1 + \log_{21}3+\log_{21} 7+\log_{21} 9 + \log_{21} 47 + \log_{21} 63 + \log_{21} 147+\log_{21} 441$. To this sum, you have have to add $\log_{21} {21}$, which is not included. – minori minus May 13 at 23:44