Solve for x log2 x + log4 x + log8 x =11/6

First, we'll impose the constraints of existence of
logarithms:

x>0

Now,
we'll change the bases of logarithms to base 2:

log2 x =
log4 x*log2 4

log4 x = log2 x/log2
(2^2)

log4 x = log2 x/2log2
2

But log2 2 = 1

log4 x = log2
x/2

log2 x = log8 x*log2
8

log8 x = log2 x/3log2 2

log8
x = log2 x/3

We'll re-write the
equation

log2 x + (log2 x)/2 + (log2 x)/3 =
11/6

We'll multiply by 6 both
sides:

6 log2 x + 3 log2 x + 2 log2 x  =
11

We'll add like
terms:

11 log2 x = 11

We'll
divide by 11:

log2 x = 1

x =
2^1

x = 2

Since 2 belongs to
the interval of admissible values, we'll validate it as solution of the
equation:

x =
2