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.

Answer: 17

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.