logx(9x^2)[log3(x)]^2=4

Question

i got the x=1/9 and x=3 is that correct?

Answer 1

Hard to say. Your syntax is murky. How did you arrive at your answer?
You appear to have log(base x) of 9x^2 or log_10(x) * 9x^2
and log(base 3) of x
And what happens when you multiply logs?
When you divide logs, you change base...

Answer 2

If the question is
log_x(9x^2)[log₃ (x)]^2=4
then x = 3 , 1/9 satisfies the equation.

As oobleck asked, how did you get that, other than by inspection.

I went as far as
(2 log3/logx + 2)(logx/log3)^2 = 4

when x = 3, the first bracket becomes (4) and the second bracket becomes 1, so it works
if x = 1/9, the first bracket becomes (1) and the 2nd becomes (-2)^2, so it works

Answer 3

Nice work, Reiny. Using u = log3/logx, we could have proceeded with
(2u+2)/u^2 = 4
u+1 = 2u^2
2u^2-u-1 = 0
(2u+1)(u-1) = 0
u = -1/2 or 1
So, that means that log3/logx = -1/2 or 1
logx/log3 = -2 or 1
log_kx = -2 or 1
x = 3^-2 or 3^1
x = 1/9 or 3