There is a unique positive real number x such that the three numbers
log82x, log4x and log2x, in that order, form a geometric progression with a positive common ratio.
The number x can be written as m/n, where m and n are relatively prime positive integers.
Find m+n.

Step-by-step explanation:

If the log82x, log4x and log2x, in that order, form a geometric progression with a positive common ratio.

Let a = log82x, b = log4x and c = log2x

If a = log8 2x; 8^a = 2x... (1)

If b = log4 x; 4^b = x ... (2)

If c = log2 x; 2^c = x...(3)

Since a, b c are in GP, then b/a = c/b

Cross multiplying:

b² = ac ...(4)  

From eqn 1, x = 8^a/2

x = 2^3a/2

x = 2^(3a-1)

From eqn 2; x = 4^b

x = 2^2b

From eqn 3: x = 2^c

Equating all the values of x, we have;

2^(3a-1) = 2^2b = 2^c

3a-1 = 2b = c

3a-1 = c and 2b = c

a = c+1/3 and b = c/2  

Substituting the value of a = c+1/3 and b = c/2 into equation 4 we have;

(c/2)² = c+1/3×c

c²/4 = c(c+1)/3

c/4 = c+1/3

Cross multiplying;

3c = 4(c+1)

3c = 4c+4

3c-4c = 4

-c = 4

c = -4  

Substituting c = -4 into equation 3 to get the value of x we have;

2^c = x

2^-4 = x

x = 1/2^4

x = 1/16    

Since the number x can be written as m/n, then x = 1/16 = m/n

This shows that m = 1, n = 16

m+n = 1+16

m+n = 17  

The required answer is 17.  

d is correct

step-by-step explanation: