Question

If $a > b > 1$ and $\frac{1}{\log_a b} + \frac{1}{\log_b a} = \sqrt{1229}$, find the value of $\frac{1}{\log_{ab} b} - \frac{1}{\log_{ab} a}$.

Answer 1

35

Answer 2

Let $y = \log_b a$. The given equation is $\frac{1}{\log_a b} + \frac{1}{\log_b a} = \sqrt{1229}$. Using the property of logarithms $\frac{1}{\log_x y} = \log_y x$, the equation becomes: $\log_b a + \log_a b = \sqrt{1229}$. Since $\log_a b = \frac{1}{\log_b a}$, we can write the equation in terms of $y$: $y + \frac{1}{y} = \sqrt{1229}$.

We need to find the value of the expression $E = \frac{1}{\log_{ab} b} - \frac{1}{\log_{ab} a}$. Using the property $\frac{1}{\log_x y} = \log_y x$, the expression becomes: $E = \log_b(ab) - \log_a(ab)$. Using the property $\log_x(yz) = \log_x y + \log_x z$, we expand the terms: $\log_b(ab) = \log_b a + \log_b b$. $\log_a(ab) = \log_a a + \log_a b$. So, $E = (\log_b a + \log_b b) - (\log_a a + \log_a b)$. Since $\log_b b = 1$ and $\log_a a = 1$, and using $y = \log_b a$ and $\log_a b = \frac{1}{y}$, we get: $E = (y + 1) - (1 + \frac{1}{y}) = y + 1 - 1 - \frac{1}{y} = y - \frac{1}{y}$.

We are given $y + \frac{1}{y} = \sqrt{1229}$ and we need to find $y - \frac{1}{y}$. We know the algebraic identity $(A - B)^2 = (A + B)^2 - 4AB$. Let $A = y$ and $B = \frac{1}{y}$. Then $AB = y \cdot \frac{1}{y} = 1$. So, $(y - \frac{1}{y})^2 = (y + \frac{1}{y})^2 - 4(y \cdot \frac{1}{y}) = (y + \frac{1}{y})^2 - 4$. Substitute the value of $y + \frac{1}{y}$: $(y - \frac{1}{y})^2 = (\sqrt{1229})^2 - 4 = 1229 - 4 = 1225$. Taking the square root of both sides: $y - \frac{1}{y} = \pm \sqrt{1225}$. We calculate $\sqrt{1225}$: $30^2 = 900$, $40^2 = 1600$. The number ends in 5, so the root must end in 5. Let's try 35. $35^2 = (30+5)^2 = 900 + 2 \times 30 \times 5 + 25 = 900 + 300 + 25 = 1225$. So, $y - \frac{1}{y} = \pm 35$.

We need to determine the sign of $y - \frac{1}{y}$. We are given that $a > b > 1$. Consider $y = \log_b a$. Since $b > 1$, the logarithm function with base $b$ is an increasing function. Since $a > b$, we have $\log_b a > \log_b b$. $\log_b b = 1$. So, $\log_b a > 1$, which means $y > 1$. If $y > 1$, then $\frac{1}{y} < 1$. Therefore, $y - \frac{1}{y}$ must be positive. So, $y - \frac{1}{y} = 35$.

The value of the expression $\frac{1}{\log_{ab} b} - \frac{1}{\log_{ab} a}$ is 35.