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 correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.

you are uyytr tyy

step-by-step explanation: