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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?

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3 Answers 3

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$$\dfrac{1}{\log_{ab}a}-\dfrac{1}{\log_{ab}b}=\log_ab\,-\log_ba \tag{1}$$

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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  • $\begingroup$ Oh , completely forgot about the thing that we get a equation like that , I just wasted my time . $\endgroup$ Commented Aug 31, 2020 at 14:09
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The formula $(x-y)^2=(x+y)^2-4xy$ gives you required expression =$sqrt(1225)$=35.

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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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