<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] Instead of trying to find the values of $\log b$ and $\log a$, just look at the equation $\log_b a + \frac 1{\log_b a} = 1229$.
Now, if $(ab)^x = a$, then $a^xb^x = a$ so $b^x = a^{1-x}$, then $a = b^{\frac x{1-x}}$, therefore $\frac x{1-x} = \log_b a$, therefore $x = \frac{\log_b a}{\log_b a + 1} = \log_{ab} a$.
Similarly, if $(ab)^y = b$, then $a^y = b^{1-y}$ so $a = b^{\frac{1-y}y}$ so $\frac {1-y}y = \log_b a$ so $y = \frac 1{\log_b a + 1} = \log_{ab} b$.

We have to find $\frac 1y - \frac 1x = {\log_b a+1} - \frac{(\log_b a + 1)}{\log_b a} = \log_b a - \frac 1{\log_b a}$.
Let $\log_b a = z$. Then $z + \frac 1z = \sqrt{1229}$, and we have to find $z- \frac 1z$. This is usual from $$
\left(z+\frac 1z\right)^2 - \left(z-\frac 1z\right)^2 = 4 \times z \times \frac 1z = 4
$$
and the fact that $a>b$ so $z=\log_b a >1$, hence $z-\frac 1z< 0$. (So the square root is the positive square root).
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