The value of $\log_a \left(\frac{a}{b}\right) + \log_b \left(\frac{b}{a}\right)$, for $1 < a \leq b$ cannot be equal to

We need to find what value the expression $\log_a \left(\frac{a}{b}\right) + \log_b \left(\frac{b}{a}\right)$ cannot equal, given that $1 < a \leq b$.

---

Let $A = \log_a \left(\frac{a}{b}\right) + \log_b \left(\frac{b}{a}\right)$

Using the logarithm property $\log_x \left(\frac{p}{q}\right) = \log_x p - \log_x q$:

$A = \log_a a - \log_a b + \log_b b - \log_b a$

---

Since $\log_x x = 1$ for any positive base $x

eq 1$:

$A = 1 - \log_a b + 1 - \log_b a$

$A = 2 - (\log_a b + \log_b a)$

---

Here's where we need a key insight! We can use the change of base formula: $\log_b a = \frac{1}{\log_a b}$

Why does this work? If we let $x = \log_a b$, then $a^x = b$. Taking $\log_b$ of both sides: $x \log_b a = 1$, so $\log_b a = \frac{1}{x} = \frac{1}{\log_a b}$.

Therefore:

$A = 2 - \left(\log_a b + \frac{1}{\log_a b}\right)$

---

Let $t = \log_a b$. Since $1 < a \leq b$, we have $t \geq 1$.

We need to find the minimum value of $t + \frac{1}{t}$ where $t \geq 1$.

Using the AM-GM inequality: For positive numbers, $\frac{t + \frac{1}{t}}{2} \geq \sqrt{t \cdot \frac{1}{t}} = 1$

This gives us: $t + \frac{1}{t} \geq 2$

When does equality occur? When $t = \frac{1}{t}$, which means $t = 1$. This happens when $a = b$.

---

Since $\log_a b + \frac{1}{\log_a b} \geq 2$:

$A = 2 - \left(\log_a b + \frac{1}{\log_a b}\right) \leq 2 - 2 = 0$

The maximum value of $A$ is $0$ (achieved when $a = b$).

As $t = \log_a b$ increases (meaning $b$ gets much larger than $a$), the expression $t + \frac{1}{t}$ increases without bound, making $A$ approach $-\infty$.

---

The range of $A$ is $(-\infty, 0]$.

Therefore, $A$ cannot equal 1 since $1 > 0$.

This problem beautifully combines logarithm properties with the AM-GM inequality. Remember that when we see expressions like $x + \frac{1}{x}$, we think AM-GM to find extrema!