Question 1174904
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Each log mentioned is base x


y = log(27/8) + log(9/4) + log(3/2) + ...  (summing 10 terms in this pattern)


y = log((3/2)^3) + log((3/2)^2) + log((3/2)^1) + ...


y = 3*log(3/2) + 2*log(3/2) + 1*log(3/2) + ...


y = (3+2+1+...)*log(3/2)


Now we must compute the arithmetic series 3+2+1+... all the way up to the tenth term of this.


We see the first term is a = 3 and the common difference is d = -1
We want to sum n = 10 terms


S = (n/2)*(2*a + d(n-1))
S = (10/2)*(2*3 - (10-1))
S = -15
The first ten terms of 3+2+1+... add up to -15


This means
y = (3+2+1+...)*log(3/2)
becomes
y = -15*log(3/2)


Then we can use the log rule log(A/B) = log(A)-log(B) to get


y = -15*log(3/2)
y = -15*(log(3) - log(2))
y = 15*(-log(3) + log(2))
y = 15*(log(2) - log(3))


Overall, this shows why 
log(27/8) + log(9/4) + log(3/2) + ...
turns into
15*(log(2) - log(3))


Each log mentioned in this solution is base x. To avoid dividing by zero, we cannot have x = 1. We also can't have {{{x <= 0}}} if we want the output of the log to be a real number. 
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