Problem For Year 2029

First simplify few things...

$(1)$ Recall $\log_a b=\dfrac{1}{\log_b a}$

& $(2)$ Substitute $\log_y x=q$ .

The given equation is:

$\large \underbrace{\log_y x+\dfrac{1}{\log_y x}}_{\large q+\frac 1q}=\sqrt{2029}$

And required expression is :

$\log_x xy-\log_y xy=(\cancel{1}+\log_x y)-(\cancel{1}+\log_y x)$

which is same as $\log_x y-\log_y x=\dfrac 1q-q$ . Hence the question translates to:

$\text{Given}: q+\dfrac 1q=\sqrt{2029},\dfrac 1q-q=\ ?$

Now we know:

$\dfrac 1q-q=-\sqrt{\left(q+\dfrac 1q\right)^2-4}=-\sqrt{(2029)-4}=-\boxed{45}$

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NOTE:- Since $x>y$ ,

$\log_y x>1\implies (\log_y x)^2>1\implies q^2>1\implies q>\frac 1q\implies \frac 1q-q<0$