There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geo

We have:  log₈(2x),  log₄(x),  and  log₂(x)  forming a geometric progression.

 

Let's write these all in terms of log₂:

 

log₄(x)  =  log₂(x) / log₂(4)  =  log₂(x) / 2

 

log₈(2x)  =  log₂(2x) / log₂(8)  =  log₂(2x) / 3  =  [ log₂(2) + log₂(x) ] / 3  =  1/3 + log₂(x) / 3

 

So, we now have:  1/3 + log₂(x) / 3,   log₂(x) / 2,   log₂(x)

 

Just to make the writing simpler, let's write  log₂(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

 

log₂(x)  =  -4   --->   x  =  2^-4  =  1/16

log₄(x)  =  log₄(1/16)  =  -2

log₈(2x)  =  log₈(1/8)  =  -1

 

The progression is:  -1, -2, and -4