SOLUTION: Find the sum of the ap: logx(27/8), + logx(9/4), + logx(3/2)...(10 terms) Answer is 15 (logx 2 − logx 3)

Click here to see ALL problems on Sequences-and-series

Question 1174904: Find the sum of the ap:
logx(27/8), + logx(9/4), + logx(3/2)...(10 terms)
Answer is 15 (logx 2 − logx 3)
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

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 if we want the output of the log to be a real number.