Question 970739
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.