Question 970739: prove that if a series of numbers are in GP,their logarithms are in AP
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! geometric progression is An = A1 * r^(n-1)
arithmetric progression is An = A1 + (n-1)*d
start with the geometric progression of An = A1 * r^(n-1)
take the log of both sides of that equation to get:
log(An) = log(A1 * r^(n-1))
since log(a*b) = log(a) + log(b), this equation becomes:
log(An) = log(A1) + log(r^(n-1))
since log(a^b) = b*log(a), this equation becomes:
log(An) = log(A1) + (n-1)*log(r)
that's an arithmetic rogression with the common difference equal to log(r).
we'll take an example:
A1 = 100
r = 1.5
n = 5
geometric progression:
An = A1 * (r^(n-1) which becomes:
A5 = 100 * 1.5^4 which becomes:
A5 = 506.25
now we want to find log(A5).
if we are correct, then log(A5) should be equal to log(506.25).
start with:
log(A1) = log(100)
r = 1.5
n = 5
use the arithmetic progression of:
log(A5) = log(A1) + (n-1)*log(r) which becomes:
log(A5) = log(100) + 4*log(1.5) which becomes:
log(A5) = log(100) + log(1.5^4) whcih becomes:
log(A5) = log(100*1.5^4) which becomes:
log(A5) = log(506.25)
A5 = 506.25
log(A5) = log(506.25)
looks like we're good.
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