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.  

The value of m+n in which m and n are relatively prime positive integers is 17.

Geometric sequence is the sequence in which the next term is obtained by multiplying the previous term with the same number for the whole series.

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 progression is,

[tex]\log_82 x,\log_4x, \log_2x[/tex]

The above numbers are in geometric progression. Thus, the ratio of first two terms will be equal to the ratio of next two terms as,

[tex]\dfrac{\log_4x}{\log_82x}=\dfrac{\log_2x}{\log_4x}\\(\log_4x)^2={\log_2x}\times{\log_82x}\\[/tex]

Using the base rule of logarithmic function,

[tex]\left(\dfrac{\log x}{\log 4}\right)^2=\dfrac{\log x}{\log2}\times\dfrac{\log2x}{\log8}\\\left(\dfrac{\log x}{\log 2^2}\right)^2=\dfrac{\log x}{\log2}\times\dfrac{\log2x}{\log2^3}[/tex]

Using the Power rule of logarithmic function,

[tex]\left(\dfrac{\log x}{2\log 2}\right)^2=\dfrac{\log x}{\log2}\times\dfrac{\log2x}{3\log2}\\\dfrac{\log x}{4}=\dfrac{\log 2+\log x}{3}\\3\log x=4(\log2x)\\\log x^3=\log(2x)^4\\x^3=16x^4\\x=\dfrac{1}{16}[/tex]

The number x can be written as

[tex]\dfrac{m}{n}[/tex]

Here, m and n are relatively prime positive integers. Thus, the value of m+n is,

[tex]m+n=1+16=17[/tex]

Hence, the value of m+n in which m and n are relatively prime positive integers is 17.

Learn more about the geometric sequence here;

https://brainly.com/question/1509142

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