Sum the following term of progression to the last term, logx, logx^2 logx^3.... logx^120
log x^n = n log x
so you have log x (1+2+3+4 ...120)
so log x times arithmetic sequence sum with a = 1, d = 1 n from 1 to 120
see
http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
To find the sum of the terms of the given logarithmic progression, we can break it down into two steps:
Step 1: Find the common ratio (r) of the progression.
Step 2: Use the formula for the sum of a geometric progression to find the sum.
Step 1: Finding the common ratio (r)
In the given progression, each term is generated by taking the logarithm of x raised to a certain power. We can see that the exponent increases by 1 for each term.
The general form of the terms in a geometric progression is a, ar, ar^2, ar^3, ..., ar^(n-1), where 'a' is the first term, 'r' is the common ratio, and 'n' is the total number of terms.
In our case, the first term is logx and the common ratio can be expressed as (logx^2 / logx), (logx^3 / logx^2), (logx^4 / logx^3), and so on.
Simplifying the common ratio expression, we have r = (logx^2 / logx) = logx.
Step 2: Finding the sum of the progression
The formula for the sum of a geometric progression is:
S = a * (r^n - 1) / (r - 1)
In this case, the first term (a) is logx, the common ratio (r) is logx, and the number of terms (n) is 120.
Using the formula, we can calculate the sum (S) of the progression:
S = logx * (logx^120 - 1) / (logx - 1)
Please note that the formula assumes that the value of logx is not equal to 1.
Now, by substituting the value of logx, you can calculate the sum of the given logarithmic progression.