There is a unique positive real number such that the three numbers , , and , in that order, form a geometric progression with positive common ratio. The number can be written as , where and are relatively prime positive integers. Find .

Guest Jun 6, 2020

#1**0 **

We have: log_{8}(2x), log_{4}(x), and log_{2}(x) forming a geometric progression.

Let's write these all in terms of log_{2}:

log_{4}(x) = log_{2}(x) / log_{2}(4) = log_{2}(x) / 2

log_{8}(2x) = log_{2}(2x) / log_{2}(8) = log_{2}(2x) / 3 = [ log_{2}(2) + log_{2}(x) ] / 3 = 1/3 + log_{2}(x) / 3

So, we now have: 1/3 + log_{2}(x) / 3, log_{2}(x) / 2, log_{2}(x)

Just to make the writing simpler, let's write log_{2}(x) as A:

---> 1/3 + A/3, A/2, A

To find the common ratio, we can divide the third term by the second term:

---> (A) / (A/2) = 2 (multiply both the numerator and denominator by 2)

We now that this ratio is also true when we divide the second term by the first term:

---> (A/2) / (1/3 + A/3) = 2

---> (3A) / (2 + 2A) = 2 (multiplying both the numer and denom by 6)

---> 3A = 2(2 + 2A) (cross multiplying)

---> 3A = 4 + 4A

---> -A = 4

---> A = -4

log_{2}(x) = -4 ---> x = 2^{-4} = 1/16

log_{4}(x) = log_{4}(1/16) = -2

log_{8}(2x) = log_{8}(1/8) = -1

The progression is: -1, -2, and -4

geno3141 Jun 7, 2020