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