If 2^(x) = 4^(y) = 8^(z) and 1/(2x) + 1/(4y) + 1/(4z) = 4 then what is the value of x?

Question

x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be

Answer 1

2^x = 4^y = 8^z ;;;; x=2y=3z;;;

[1/2x] + [1/4y] + [1/4z] = 4

multiply 4 both sides ;;

[2/x] + [1/y] + [1/z] = 16 ;;;

[2/x] + [2/x] + [3/x] = 16 ;;;

[7/x] = 16 ;;;

x= 7/16 ;;;

Answer 2

2^x=4^y=8^z......................................................given
2^x=2^(2y)=2^(3z)
x=2y=3z

x.............................................................................(i)
y=x/2.....................................................................(ii)
z=x/3....................................................................(iii)

[1/2x+1/4y+1/4z]..............................................given
{(1/2x)+[1/4(x/2)]+[1/4(x/3)]........................from (i),(ii),(iii)
Simplify
x=7/16
Ans is 7/16

Answer 3

we have
2^(x) = 4^(y) = 8^(z)
or, 2^(x) = 2^(2y) = 2^(3z)
-> x=2y=3z

also we have
1/(2x) + 1/(4y) + 1/(4z) = 4
1/(2x) + 1/(2x) + 1/(4x/3) = 4
1/(2x) + 1/(2x) + 3/(4x) = 4
or, 7/4x=4
16x=7
hence,
x=7/16