how can i evaluate logarithmic series

Question

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 '19 at 23:12

score 2 · Accepted Answer · answered May 13 '19 at 23:15

2

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 $$\log_{21} {1} + \log_{21} {441} = \log_{21} {3} + \log_{21}{\frac{441}3} = \log_{21} {7} + \log_{21}{\frac{441}7} = \text{ ... }= 2$$

answered May 13 '19 at 23:15

minori minus

91

so would the answer be 21^2? – user477465 May 13 '19 at 23:18
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 '19 at 23:22
Also note that only $21$ doesn't have a pair. – minori minus May 13 '19 at 23:23
but would the number of pairs be 220? so it would just be 220*2? – user477465 May 13 '19 at 23:38
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 '19 at 23:44
ah gotcha, so the answer should be 9? – user477465 May 14 '19 at 00:39
Yes. $\text{ }$ – minori minus May 14 '19 at 00:41

how can i evaluate logarithmic series

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