can someone explaine to me how to find the x?

I think this is meant to be:

\(\log_2x=\log_4{4x}\)

Let:

 \(a=\log_2x\\x=2^a\)

So, also:

\(a=\log_4{4x}\\4x=4^a\\x=4^{a-1}\\x=2^{2(a-1)}\)

Hence we must have:

a = 2(a - 1)   so that  a = 2, which means x = 2^a  or x = 4 

.

Solve for x:
(log(x))/(log(2))=(log(4 x))/(log(4))

Subtract (log(4 x))/(log(4)) from both sides:
(log(x))/(log(2))-(log(4 x))/(log(4))=0

Bring (log(x))/(log(2))-(log(4 x))/(log(4)) together using the common denominator log(2) log(4):
-(log(2) log(4 x)-log(4) log(x))/(log(2) log(4))=0

Divide both sides by -1/(log(2) log(4)):
log(2) log(4 x)-log(4) log(x)=0

log(2) log(4 x)-log(4) log(x)=log(x^(-2 log(2)))+log(4^(log(2)) x^(log(2)))=log(4^(log(2)) x^(-log(2))):
log(4^(log(2)) x^(-log(2)))=0

Cancel logarithms by taking exp of both sides:
4^(log(2)) x^(-log(2))=1

Divide both sides by 4^(log(2)):
x^(-log(2))=4^(-log(2))

Take reciporicals of both sides:
x^(log(2))=4^(log(2))

Raise both sides to the power of 1/(log(2)):
Answer: |  x=4

Why are the 2 and the 4 written as strange superscripts ?

Our guest seems to understand but I don't   

Here's another approach....using Alan's interpretation

log₂x = log₄(4x)

log₂x  = log₄4  + log₄x

log₂x - log₄x  = log₄4

log₂x - log₄x  = 1

logx/log2  - logx/log 4   = 1

logx [ log4 - log 2] / [log2 * log 4]  = 1

logx  = [log2 * log4] / [ log4 - log2]

logx = [log2 * log4] / [log (4/2)]

logx = [log2 * log4] / [log2]

log x  = log 4

x = 4