[Riley_5000]

I remember seeing a ln paradoxical proof a while ago. I can't remember it, so I came up with my own (I hope it's not the same!!)

say x =1

ln ((x^-2)/(x^2)) = 0

lnx^-2 - lnx^2 =0

lnx^-2 = lnx^2

-2lnx = 2lnx

-2 = 2

What do you think?

[azmanam]

(x^-2/x²) = 1/x⁴.  It does not equal 1.

[P]

(x^-2/x²) = 1/x⁴.  It does not equal 1.

does when x = 1 though!   

[azmanam]

...

which is what op stipulated in his first statement...

no math in the morning for me...

carry on.

[Yggdrasil]

As with all contradictory proofs, it involves division by zero.  ln(x)=ln(1)=0.

[Borek]

As with all contradictory proofs, it involves division by zero.

Not all, I have seen much more subtle one, using complex numbers

[Riley_5000]

Substitute "n" for "-2"

you get :
-n=n

It doesn't matter what x equals! Why do I do this to myself?

[Yggdrasil]

ln ((x^-2)/(x^2)) = 0 is true only for x=1. 

$$ log \left( \frac{(1/4)}{4} \right) = log(1/16) \neq 0 /$$

[vhpk]

I remember seeing a ln paradoxical proof a while ago. I can't remember it, so I came up with my own (I hope it's not the same!!)

say x =1

ln ((x^-2)/(x^2)) = 0

lnx^-2 - lnx^2 =0

lnx^-2 = lnx^2

-2lnx = 2lnx

-2 = 2

What do you think?

All of your expression are trues except the final step. Since lnx = 0 as x is equal to 1, you divide lnx for both sides i.e you divide 0 -> that's the mistake you've made.
Good luck