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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 .

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 Jun 6, 2020
 #1
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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

 Jun 7, 2020

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