How to work with sums? $\sum$ During high school I missed out on some of the math lessons due to health issues, one thing I missed out on was working with sums. I am now a second-year engineer student and I am still amazed how bad I am at this. I am reading a math course right now (statistics) and finally realized how big this issue is so here I am for help.
let's say we got a sample of x:
$x_1,x_2,...,x_9$
and we got these two equations:
$$\sum_{i=1}^9(ln(x_i)+ln(h))$$
$$\sum_{i=1}^9(ln(x_i)-ln(h))$$
h is some variable. How do we "break out" ln(h) of the sum so only
$$\sum_{i=1}^9(ln(x_i))$$
is within the sum. I think it is possible to multiply ln(h) with 9 and put it infront like this:
$$9ln(h)\sum_{i=1}^9(ln(x_i))$$ but I am not sure if it is correct and how the seccond equation would become? Is there any website with rules when working with this kind of summation?
 A: $$\sum_{i=1}^9(\ln x_i+\ln h)$$
It is not permitted to multiply on the left by $\ln h$, but it is permitted to put the $\ln h$ terms in a separate sum:
$$=\sum_{i=1}^9\ln x+\sum_{i=1}^9\ln h$$
Since this new sum's terms do not depend on the changing variable we can replace it by (the term) times (the number of indexes):
$$=\sum_{i=1}^9\ln x+9\ln h$$
Similar reasoning applies for the sum with $-$ in it.
A: If you're confused about how to use shorthand notation like this, take a step back and consider what it is a shorthand for. In this case, $\sum_{i=1}^9(\ln(x_i)+\ln(h))$ is shorthand for
$$
(\ln(x_1)+\ln(h)) + (\ln(x_2)+\ln(h)) + (\ln(x_3)+\ln(h)) + \\
(\ln(x_4)+\ln(h)) + (\ln(x_5)+\ln(h)) + (\ln(x_6)+\ln(h)) + \\
(\ln(x_7)+\ln(h)) + (\ln(x_8)+\ln(h)) + (\ln(x_9)+\ln(h))
$$
Once we remove the (now unnecessary) parentheses, we see that we have nine copies of $\ln(h)$ added together, and the standard shorthand for that is $9\cdot \ln(h)$. So using this and reorganising the terms a bit we get
$$
9\cdot \ln(h) + \ln(x_1) + \ln(x_2) + \ln(x_3) + \ln(x_4) + \\\ln(x_5) + \ln(x_6) + \ln(x_7) + \ln(x_8) + \ln(x_9)
$$
And by now you can hopefully see that apart from the first term, all the other terms may be collected to become $\sum_{i = 1}^9\ln(x_i)$. So we get our answer
$$
\sum_{i = 1}^9(\ln(x_i)+\ln(h)) = 9\cdot \ln(h) + \sum_{i = 1}^p\ln(x_i)
$$
A: You should first convince yourself that you can "take out" $\ln(h)$ of both your numbers:
$$\sum_{i=1}^9 (\ln(x_i) + \ln(h)) = 9\ln(h) + \sum_{i=1}^9 \ln(x_i)$$
You can also do it for the second number (be careful with the sign).
Maybe you could try to do some exercises from this here:
https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/summationdirectory/Summation.html
