Logarithm Question can someone help me?

Split the 9 into 3 * 3:
(3*3)^log_3(log x) = log(x) - 2 log²(x) + 4

Use this rule --> (a*b)^c = a^c * b^c:
3^log_3(log x) * 3^log_3(log x) = log(x) - 2 log²(x) + 4

Simplify the two logs on the left:
log(x) * log(x) = log(x) - 2 log²(x) + 4
log²(x) = log(x) - 2 log²(x) + 4

Get everything on the left hand side:
3 log²(x) - log(x) - 4 = 0

Use this substitution --> u = log(x)
3u² - u - 4 = 0

Factor:
(u + 1)(3u - 4) = 0

Find the two possible values:
u = -1
or
u = 4/3

Substitute back in for u:
log(x) = -1
or
log(x) = 4/3

But we can't take log_3 of a negative number, so we can drop the negative answer.

Next, I'm not sure if you are using 'log' to mean the common log (base 10) or the natural log (base e). 

Assuming the common log, raise both sides upon the base 10:
x = 10^(4/3)
x ≈ 21.5443469...

(If it is the natural log, then x = e^(4/3) ≈ 3.79366789...)

The answer is as follows:

You cant show that they are the same because they (generally) are not.
Plug some values  for x and evaluate both sides. You will see that the differ. 

Like, if x = 1000, what do you get both sides?
If x = 1, what do you get both sides?