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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

1. $– 0.5$
2. $1$
3. $0$
4. $– 1$

Given that, $1 < a \leq b$

Now, $\log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = \log_{a}a – \log_{a}b + \log_{b}b – \log_{b}a$

$\Rightarrow \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = 1 – \log_{a}b + 1 – \log_{b}a$

$\Rightarrow \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = 2 – \left( \log_{a}b + \log_{b}a \right)$

$\Rightarrow \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = 2 – \left( \log_{a}b +\frac{1}{\log_{a}b} \right) \quad \longrightarrow (1)$

Now, using $\text{AM} \geq \text{GM}$

$\frac{ \log_{a}b + \frac{1}{\log_{a}b}} {2} \geq \sqrt{ \log_{a}b \cdot \frac{1}{\log_{a}b}}$

$\Rightarrow \log_{a}b + \frac{1}{\log_{a}b} \geq 2$

From equation $(1),$

$\log_{a} \left( \frac{a}{b} \right) + \log_{b} \left( \frac{b}{a} \right) = \underbrace{2 – \underbrace{\left( \log_{a}b + \frac{1}{\log_{a}b} \right)}_{\geq 2}}_{\leq 0}$

Thus, it can never be equal to $1.$

Correct Answer$: \text{B}$

$\textbf{PS:}$ Important logarithm properties :

• $\log_{a}a = 1$
• $\log_{a} \left( \frac{m}{n} \right) = \log_{a}m – \log_{a}n$
• $\log_{a} (mn) = \log_{a}m + \log_{a}n$
• $\log_{a^{x}}b = \frac{1}{x} \log_{a}b$
• $\log_{a}b = \frac{1}{\log_{b}a}$
• $\log_{a}b = x \Rightarrow b = a^{x}$
• $\log_{a}b = \frac{\log_{c}b}{\log_{c}a}$
• $\log_{a}1 = 0$
• $\log_{b}a^{x} = x \log_{b}a$
• $a^{\log_{a}b} = b$
• $\log_{a}^{k} n = \left( \log_{a}n \right)^{k}$
• $a^{\log_{b}{x}} = x^{\log_{b}{a}}$
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