<s>[INST] If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$ 
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$ .

What I Tried :- I tried the problem this way :-
As $log_a(b) = \frac{log_b(b)}{log_b(a)}$ , we have $\frac{1}{log_a(b)} = \frac{log_b(a)}{log_b(b)} = log_b(a).$ So :-
$$log_b(a) + \frac{1}{log_b(a)} = \sqrt{1229}$$
$$\rightarrow \frac{log(a)}{log(b)} + \frac{log(b)}{log(a)} = \sqrt{1229}$$
$$\rightarrow \frac{(log(a))^2+(log(b))^2}{log(a)log(b)} = \sqrt{1229}$$
Now :-
$$\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$$
$$\rightarrow \frac{log(a) + log(b)}{log(b)} - \frac{log(a) + log(b)}{log(a)}$$
$$\rightarrow \frac{(log(a))^2 + log(a)log(b) - log(a)log(b) - (log(b))^2}{log(a)log(b)}$$
$$\rightarrow \frac{(log(a))^2 - (log(b))^2}{log(a)log(b)}$$
$$\rightarrow \sqrt{1229} - \frac{2(log(b))^2}{log(a)log(b)}$$
I could conclude only upto this, other than that I have no idea . Now can anyone help me?
 [/INST] $$\dfrac{1}{\log_{ab}a}-\dfrac{1}{\log_{ab}b}=\log_ab\,-\log_ba \tag{1}$$
Let $\log_b a=x$, we get the first expression to be $$x +\frac{1}{x}=\sqrt{1229} \tag{2}$$,
Now, squaring (1) and (2), we see (1) becomes:
$$x^2+\frac{1}{x^2} + 2 = 1229\tag{3}$$
and expression (2) becomes:
$$x^2-2+\frac{1}{x^2} = A^2\tag{4}$$
Subtracting (3) and (4), we get:
$$1229-A^2=4$$
Therefore, A=$\sqrt{1225}$
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